Markov Processes, Brownian Motion, and Time Symmetry, Second Edition (Grundlehren der mathematischen Wissenschaften)

E

iC, we have Pt
~
as t

t 0,

and

vanishes only at 8, like <po Write

then {Zt, ~} is a positive supermartingale by Proposition 2 of §2.1, and {Zt} has right continuous paths with left limits, because {Xt} does and U 1

°

((w) = inf{t > 0IZt-(w) =

°

°

or Zt(w) = O},

and the theorem follows from Theorem 4 of §1.5.

o

Corollary. If (Pt) is strictly Markovian, then almost surely the sample function is bounded in each finite t-interval. Namely not only X(t, w) E E for all t ;::: 0, but X (t - , w) E E for all t > as well.

°

Proof. Under the hypothesis we have

p{X t

E

E for all t E Tn Q} = 1,

It follows from the theorem that for each t

P{( <

Hence P{( < <Xl}

=

E

Q:

t} = P{( < t; XtEE} = 0.

0, which is equivalent to the assertion ofthe Corollary.

o

Theorem 7 is a case of our discussion is §1.2 regarding 8 as an absorbing state: cf. (13) there. Exercises 1. Consider the Markov chain in Example 1 of §1.2, with E the set of positive integers, no 8, and Iff the discrete topology on E. Show that iflimqo Pii(t) = I uniformly in all i, then property (I) is true for every f Take a version which is right continuous and has left limits. Show that for each sample function

2. Basic Properties

56

X(',w), there is a discrete set oft at which X(t-,w) =f. X(t,w), and X(·,w) is constant between two consecutive values of t in this set. 2. Show that the semigroups of Examples 2, 3 and 4 in §1.2 are all Fellerian. 3. For a Markov chain suppose that there exists an i such that lim qo [1 Pii(t)]/t = + 00. (Such a state is called instantaneous. It is not particularly easy to construct such an example; see Chung [2J, p. 285.) Show that for any fJ > 0 we have pi{X(t) = i for all t E [O,fJJ} = O. Thus under pi, no version of the process can be right continuous at t = O. Hence (Pt) cannot be Fellerian. Prove the last assertion analytically. 4. Prove that in the definition of Feller property, condition (2) implies condition (1). [Hint: by the Riesz representation theorem the dual space of C is the space of finite measures on Ea. Use this and the Hahn-Banach thorem to show that the set of functions of the form {UaqJ}, where a > 0 and qJ E C, is dense in C. If f belongs to this set, (1) is true.J

2.3. Strong Markov Property and Right Continuity of Fields We proceed to derive several fundamental properties of a Feller process. It is assumed in the following theorems that the sample functions are right continuous. The existence of left limits will not be needed until the next section. We continue to adopt the convention in §1.3 that X 00 = 0, which makes it possible to write certain formulas without restriction. But such conventions should be watched carefully. Theorem 1. For each optional T, we have for each fEe, and u :2: 0:

(1)

Proof. Observe first that on {T = oo}, both sides of (1) reduce to f(o). Let T

=

n

[2 TJ

2n

n

+1 '

so that Tn H T, each Tn is strictly optional, and takes values in the dyadic set D (see (19) of §1.3). Recall from Theorem 5 of §1.3 that 00

ffT+

=

1\

ffTn ·

n=1

Hence for any A E ffT +, if we write Ad = A n {Tn = d} for dE D, then ffd. Now the Markov property applied at t = d yields

Ad E

2.3. Strong Markov Property and Right Continuity of Fields

57

see (2) of §1.2. Since Tn = d on Ad' we have by enumerating the possible values of Tn:

Since both! and Ptf are bounded continuous, and X(·) is right continuous, we obtain by letting n --+ 00 in the first and last members of (2):

L

!(XT+JdP

The truth ofthis for each A

E ff'T+

=

L

Pu!(XT)dP.

is equivalent to the assertion in (1).

0

When T is a constant t, the equation (1) reduces to the form of Markov property given as (iic) in §1.1, with the small but important improvement that the:!i';; there has now become :!i';;+. The same arguments used in §1.1 to establish the equivalence of (iic) with (iib) now show that (l)remains true for ! E btff. More generally, let us define the "post- T field" ff'~ as follows: ff'~

=

a(X T+u' U :2: 0);

which reduces to ff'; when T = t. Then we have for each integrable Y in

ff'~:

This is the extension of (iia) in §1.1. Alternatively, we have for each integrable Yin ffo: (3) Definition. The Markov process {Xt,:!i';;} is said to have the strong Markov property iff (3) is true for each optional time T.

Thus Theorem 1 is equivalent to the assertion that: A Feller process with right continuous paths has the strong Markov property. An immediate consequence will be stated as a separate theorem, which contains the preceding statement when T == O. Theorem 2. For each optional T, the process {X T + t' ff'T + t +, t E T} is a Markov process with (Pt) as transition semigroup. Furthermore, it has the strong Markov property.

58

2. Basic Properties

Proof. Recall that T + t is optional, in fact strictly so for t > 0; and X T+t Y7T+t+ by Theorem 10 of §1.3. Now apply (1) with T replaced by T + t:

E

(4)

This proves the first assertion of the theorem. As for the second, it is a consequence of the re-statement of Theorem 1 because {X T+P Y7T+t +} is a Feller process with right continuous paths. 0 There is a further extension of Theorem 1 which is useful in many applications. Indeed, the oldest case of the strong Markov property, known as Desire Andre's reflection principle, requires such an extension (see Exercise 12 of §4.2).

Theorem 3. Let S ~ T and S E Y7T+' Then we have for each f

E

C: (5)

where S - T is defined to be 00

if S =

00

(even if T = (0); and X co = o,f(o) = O.

The quantity in the right member of (5) is the value of the function (t, x) -> W{j(X t)} = PJ(x) at (S(w) - T(w), XT(w)), with w omitted. It may be written out more explicitly as

fQ f(X

S(w) - T(W)(W')

)pxT(W)(dw').

(6)

It is tempting to denote the right member of(5) by P s - T f(X T) in analogy with (1), but such a notation would conflict with our later, and established, usage of the balayage operator P T in §3.4.

Proof of Theorem 3. Observe first that (5) reduces to f(o) = f(o) on {S = oo} by the various conventions, so we may suppose S < 00 in what follows. Let

then Un 1 (S - T). Since Sand T belong to Y7T +, so does Un for each n. Consequently, for each d ED (the dyadic set) we have {Un = d} E Y7T +. Now let A be any set in Y7T+, and put Ad = A n {Un = d} E Y7T+. We have for each f E iC:

2.3. Strong Markov Property and Right Continuity of Fields

59

where the second equation is by (1). Now t ~ Ptf(x) is continuous for each x and fEe by Theorem 1 of §2.2. Letting n ~ 00 in the first and last members of (7), we obtain by right continuity:

D

This is equivalent to (5).

The next theorem demands a thorough understanding of the concepts of "completion" and "augmentation". For the first notion see Theorem 2.2.5 of the Course. The second notion will now be discussed in detail. Let (Q, 5', P) be a complete probability space. The class of sets C in 5' with P( C) = 0 will be denoted by % and called the P-null sets. Completeness means: a subset of a P-null set is a P-null set. Let f§ be a sub-(}-field of 5'. Put f§*

= (}(f§ v %),

namely f§* is the minimal (}-field including f§ and %. f§* is called the augmentation of f§ with respect to (Q, 5', P). It is characterized in the following proposition, in which a "function" is from Q into [ - 00, + 00]' Recall that for a function Yon E, Y E f§* is an abbreviation of Y E f§* II! where I! is the Borel field of E.

Lemma. A subset A of Q belongs to f§* if and only if there exists B E f§ such that A 6. B E %. A function Y belongs to f§* if and only if there exists a function Z E f§ such that {Y # Z} E %. Proof. Let d denote the class of sets A described in the lemma. Observe that A 6. B = C is equivalent to A = B 6. C for arbitrary sets A, Band C. Since f§* contains all sets of the form B 6. C where BE f§ and C E %, it is clear that f§* ~ d. To show that s/ ~ f§* it is sufficient to verify that d is a (}-field since it obviously includes both f§ and %. Since AC Ll BC= A Ll B, d is closed under complementation. It is also closed under countable union by the inclusion relation

and the remark that if the right member above is in %, then so is the left member. Thus d is a (}-field and we have proved the first sentence of the Lemma. Next, let Z E f§ and {Y # Z} E %. Then for any subset S of [ - 00, + 00], we have the trivial inclusion

{Y E S} 6. {Z E S}

c

{Y # Z}.

60

2. Basic Properties

For each Borelian S, we have {Z E S} E rJ; hence {Y E S} E rJ* by what we have proved, and so Y E rJ*. Conversely, let Y E rJ*, and Y be countablyvalued. Namely Y is of the form I.f=l C)Aj' where the c/s are points in [ - 00, + 00], and the A/s are disjoint sets in rJ*. Then for eachj there exists B j E rJ such that A j 6. B j E fl. Since the A/s are disjoint, B; 1\ B j E fl if i;6 j. Put B /l = B l , and Bj = B1 ... Bj-l B j forj ~ 2, then B j 6. Bj Efland so A j 6. Bj E fl. The Bj's are disjoint. Put Z = I.f= 1 c) Bj' Then Z E rJ and 00

{Y;6 Z}

E

U {A

j

6. Bj}

E

fl.

j= 1

Thus Y is of the form described in the second sentence of the lemma. If Y E rJ* there exists countably valued y(k) E rJ* such that y(k) --+ Y. Let Z(k) E rJ such that {y
{Y;6 Z} c

U {y(k);6 Z(k)} E fl.

o

k= 1

Remark. If (Q,:Ii', P) is not complete, the Lemma is true provided that we assume Y E :Ii' in the second assertion. We leave this to the interested reader. Corollary. (rJ*)*

= rJ*. (Q, rJ*, P) is complete.

We now review the definition of a conditional expectation, although this has been used hundreds of times above. Let Y E :Ii' and E(I YJ) < 00. Then the conditional expectations E( Y I rJ) is the class of functions Z having the following properties: (a) (b)

Z E rJ; for each WE brJ, we have E(WZ)

= E(WY).

(8)

Note that (8) is then also true for WE brJ*. It follows that if Z is a member of the class E(YlrJ), and Z* is a member of the class E(YlrJ*), then {Z;6 Z*} E fl. We state this as follows: E(Y IrJ)

== E(Y IrJ*),

(9)

mod fl.

We are ready to return to (1) above. First put T

== t, then shrink

~+

to

:Ii'~+ as we may:

(10)

2.3. Strong Markov Property and Right Continuity of Fields

61

We have seen that there is advantage in augmenting the (I-fields ~f+ appearing in (10). Since ~f+ :::::J ~f :::::J ~g, it is sufficient to augment ~g to achieve this, for then JV will be included in it, and a fortiori in the others. Theorem 4. Suppose that ~g is augmented, then the family {~f, t E T} is right continuous.

Proof. According to (10), its right member, say Z 1, is a representative of its left member. If Z2 is any other representative of the latter, then {Zl =I Z2} E JV. This is a consequence of the definition of the conditional expectation which implies its uniqueness mod JV. Hence Z2 E ~f by the Lemma. Thus the entire class in the left member of (10) belongs to ~f. Now the same argument as used in §1.l to show that (iic) implies (iia) there yields, for any YEb~;:

(11) Let Yo E b~f, then a representative of E{Yol~f+} is Yo itself because ~f c ~f+. It follows as before that

(12) Lemma 2 of §1.l may be used to show that the class of sets Y satisfying (11) constitutes a (I-field. We have just seen that this class includes ~f and ~;. Since ~o = (I(~f, ~;) for each t, the class must also include ~o. Hence (11) is true for each Y E ~o. For any Y E ~f+ (c ~o), a representative of E{YI~f+} is Y itself. Hence Y E ~f by (11), and this means ~f+ c ~f. Therefore ~f+ = ~f as asserted. 0 According to the discussion in §2.2, Theorem 4 may be read as follows. For each probability measure J1 on g, let (Q, ~Il, PIl) be the completion of (Q, ~o, PIl); and let ~~ denote the augmentation of ~f with respect to (Q, ~Il, PIl). [Note that if Y E ~Il, then EIl(Y) is defined as E'"(Z) where Z E ~o and PIl{ Y =I Z} = 0.] Then {~~, t E T} is right continuous. Furthermore, we have as a duplicate of (10):

and more generally for Y

E

b~o, and each T optional relative to {~n:

(13)

cf. (3) above. The notation Ell indicates that the conditional expectation is with respect to (Q, ~Il, PIl); and any two representatives of the left number in

62

2. Basic Properties

(13) differs by a PU-null function. It will be shown below (in the proof of Theorem 5) that (13) is also true for Y E b~'". Now if Y E ~o, then Yo 8 t E ~o. But if Y E ~'", we cannot conclude that Yo 8 t E ~'", even if Y = f(X u )' This difficulty will be resolved by the next step. Namely, we introduce a smaller a-field (14) and correspondingly for each t ;;::: 0: (15) In both (14) and (15), 11 ranges over all finite measures on e. It is easy to see that the result is the same if 11 ranges over all a-finite measures on e. Leaving aside a string of questions regarding these fields (see Exercises), we state the useful result as follows. Corollary to Theorem 4. The family as the family {~,:, t E T} for each 11.

{~;,

t E T} is right continuous, as well

Proof. We have already proved the right continuity of {~n and will use this to prove that of {~n. Let 8(11, s) be a class of sets in any space, for each 11 and s in any index sets. Then the following is a trivial set-theoretic identity: /\ /\ 8(11, s) = /\ /\ 8(11, s). '"

Applying this to 8(11, s) = we obtain

s

s

~~,

(16)

'"

for all finite measures 11 on e, and all s E (t, (0),

/\ (~n+

= (~;)+

(17)

'"

where ( ~'") t +

= /\ ~'"

S'

s>t

(~;)+ = /\ ~;. s>t

These are to be distinguished from (~+)'" and (~+r = /\'"(~+)'", but see Exercise 6 below. Since (~n+ = ~,:, (17) reduces to~; = (~;)+ as asserted.

D After this travail, we can now extend (13) to Y in b~-. But first we must consider the function x ---+ EX {Y}. We know this is a Borel function if Y E b~o ; what can we say if Y E b~-? The reader will do well to hark back to the theory of Lebesgue measure in Euclidean spaces for a clue. There we may begin with the a-field f!4 of classical Borel sets, and then complete f!4 with

63

2.3. Strong Markov Property and Right Continuity of Fields

respect to the Lebesgue measure. But we can also complete :!Jj with respect to other measures defined on !!lJ. In an analogous manner, beginning with :F 0 , we have completed it with respect to PJl and called the result :F Jl . Then we have taken the intersection of all these completions and called it :F-. Exactly the same procedure can be used to complete Cfa (the Borel field of Eo) with respect to any finite measure f.1 defined on It. The result will be denoted by It Jl; we then put

(18) As before, we may regard CfJl and Cf as classes of functions defined on E, as well as subsets of E which will be identified with their indicators. A function in Cfll is called f.1-measurable; a function in Cf- is called universally measurable. A function f is f.1-null iff f.1( U #- O}) = O. See Exercise 3 below for an instructive example. The next theorem puts together the various extensions we had in mind. Note that :Fj = (:Fj)+ as a consequence of the Corollary to Theorem 4. Theorem 5. If Y E b:F-, then the function qJ defined on E by

(19) is in Cf-. For each f.1 and each T optional relative to {:Fn, we have (20)

(21) Proof. Observe first that for each x, since Y E :FEx , EX(y) is defined. Next, it follows easily from the definition of the completion :F Jl that Y E b:FJl if and only if there exist Y1 and Yz in b:F° such that PIl{Y1 #- Yz} = 0 and Y1 :::;; Y:::;;

Yz . Hence if we put i

=

1,2;

qJl and qJz are in Cf, qJl :::;; qJ :::;; qJz and f.1({qJl #- qJz}) = O. Thus qJ E CfJl by definition. This being true for each f.1, we have qJ E Cf-. Next, define a measure v on Cfa as follows, for each f E bCf:

This is the distribution of X T when f.1 is the distribution of X 0. Thus for any Z E b:F°, we have (22)

64

2. Basic Properties

If Y E :F-, then there exists Z 1 and Z z in :F 0 such that Pv {Z 1 #- Z z} = 0 and Zl S Y s Zz. We have Zi ()T E:F1l by (22) of §1.3, with {:Fn for {~} (recall :F1l = :F~), and 0

(23)

by (22) above. This and (24) imply (20). Furthermore, we have by (24):

Now put l/Ji(X) = EX(Z;), i = 1,2. Since Zi E :F 0, l/J;(X T) is a representative of EIl{Zi 0 ()T!:Fj} by (13). On the other hand, we have l/J1 s cp s l/Jz, hence (26) and

Comparing (25) and (26) we conclude that cp(XT) is a representative of E{Y 0 ()TI:Fj}. This is the assertion (21). D We end this section with a simple but important result which is often called Blumenthal's zero-or-one law. Theorem 6.

Let A

E

:F;; then for each x we have PX(A) = 0 or PX(A) = 1.

Proof. Suppose first that A PX{X o = x} = 1, we have

E

:Fg. Then A =

Xo:

1

(A) for some A

E

rff. Since

which can only take the value 0 or 1. If A E ~;, then for each x, there exists AX in $'8 such that PX(A 6 AX) = 0, so that PX(A) = PX(A X) which is either 0 or 1 as just proved. 0 If we think that :Fg is a small a-field, and :F; is only "trivially" larger, the result appears to be innocuous. Actually it receives its strength from the fact that ~; = (~-)o+, as part of the Corollary to Theorem 4. Whether an event will occur "instantly" may be portentous. For instance, let Tbe optional relative to {:F;-, t E T}, then {T = O} E:F; and consequently the event {T = O} has probability 0 or 1 under each PX. When T is the hitting time

2.3. Strong Markov Property and Right Continuity of Fields

65

of a set, this dichotomy leads to a basic notion in potential theory (see §3.4)the regularity of a point for a set or the thinness of a set at a point. For a Brownian motion on the line, it yields an instant proof that almost every sample function starting at 0 must change sign infinitely many times in any small time interval. Exercises 1. Prove that if \it :::::: 0: ~ = ~+, then for any T which is optional relative to {~}, we have \it:::::: 0: JFT + t = JFT + t +. 2: Let N* denote the class of all sets in JF o such that PX(A) = 0 for every x E E. Let JF* = a(JFo v N*). Show that JF* c JF-. The a-field JF* is not to be confused with JF-; see Problem 3. 3. Let Q = R 1 and E = R 1 . Define X/w) = w + t; then {Xt, t:::::: O} is the uniform motion (Example 2 of §1.2). Show that JF? = fJ1Jl, the usual Borel field on Rl, and p x = Cx' the point mass at x. Show that

/\ (JFo)"x

is the class of all subsets of R 1 ;

xeRl

/\ (JF o)'" is the class of universally measurable sets of R 1 /l where J1 ranges over all probability measures on [J#l. The only member of class N* defined in Problem 2 is the empty set so that JF*= [J#l. (This example is due to Getoor.)

4. Show that

Is it true that

5. For each n let '!J n be a class of positive (:::::: 0) functions closed under the operation of taking the lim sup of a sequence; let yt' be a class of positive functions closed under the operation of taking the lim inf of a sequence. Denote by '!J n + yt' the class of functions of the form gn + h where gn E '!J n and hE yt'. Suppose that '!In:::J '!I n+ 1 for every nand '!J = n;:'=l '!J w Prove that

[Hint: if f belongs to the class in the left member, show that f = lim sup gn + lim inf hn where gn E '!J »' hn E yt'; show that lim sup gn E '!J.] 6. Apply Problem 5 to prove that

/\ (JF?)'" = (/\ JF?)/l; s>t

s>t

66

2. Basic Properties

namely that the two operations of "taking the intersection over (t, (0)" and "augmentation with respect to pI"" performed on {3b~, sET} are commutative. 7. If both Sand T are optional relative to {3b;}, then so is T [Cf. Theorem 11 of §1.3.]

+S

0

8r .

2.4. Moderate Markov Property and Quasi Left Continuity From now on we write:#'; for 3b; and 3b for 3b-, and consider the Feller process (X t ,:#,;, pI") in the probability space (fl, 3b' pI") for an arbitrary fixed probability measure J1 on Iff. We shall omit the superscript J1 in what follows unless the occasion requires the explicit use. For each OJ in fl, the sample function X(', OJ) is right continuous in T = [0, (0) and has left limits in (0, (0). The family {3bp t E T} is right continuous; hence each optional time T is strictly optional and the pre- T field is 3br ( = 3br +). The process has the strong Marov property. Recall that the latter is a consequence of the right continuity of paths without the intervention of left limits. We now proceed to investigate the left limits. We begin with a lemma separated out for its general utility, and valid in any space (fl, 3b, P). Lemma 1. Let ';§ be a sub-(J-field of 3b and X each f E iC o we have

E ';§;

Y

E

3b. Suppose that for

E {J( Y) I';§} = f(X). Then P{X

(1)

= Y} = 1.

Proof. It follows from (1) that for each open set U:

(2)

see Lemma 1 of §1.1. Take A = {X ¢ U}, and integrate (2) over A: P{Y E U; X ¢ U}

= P{X E U;

X ¢ U}

=

o.

(3)

Let {Un} be a countable base of the topology; then we have by (3): P{X i= Y} ~

I

P{Y E Un; X ¢ Un} = 0.

D

n

Corollary. Let X and Y be two random variables such that for any f 9 E iCc we have E{J(X)g(X)}

Then P{X

= Y} = 1.

=

E{J(y)g(X)}.

E

iCc> (4)

67

2.4. Moderate Markov Property and Quasi Left Continuity

Proof. As before (4) is true if g = lA, where A is any open set. Then it is also true if A is any Borel set, by Lemma 2 of §1.1. Thus we have

and consequently

f(X) = EU(X)I~} = EU(Y)I~}, where ~ = O"(X), the O"-field generated by X. Therefore the corollary follows from Lemma 1. 0 The next lemma is also general, and will acquire further significance shortly.

Lemma 2. Let {X t , t E R} be an arbitrary stochastic process which is stochastically continuous. If almost every sample function has right and left limits everywhere in R, then for each t E R, X is continuous at t almost surely. Proof. The assertion means that for each t,

p{lim X s = Xt} = 1. s-+t

Since X is stochastically continuous at t, there exist that almost surely we have

Sn

ii t and

t n II t such

lim X Sn = X t = lim X tn . n

n

But the limits above are respectively X t - and X t + since the latter are assumed to exist almost surely. Hence P{ X t - = X t = X t +} = 1 which is the assertion.

o

Remark. It is sufficient to assume that X t - and X t + exist almost surely, for each t. The property we have just proved is sometimes referred to as follows: "the process has no fixed time of discontinuity." This implies stochastic continuity but is not implied by it. For example, the Markov chain in Example 1 of §1.2, satisfying condition (d) and with ajinite state space E, has this property. But it may not if E is infinite (when there are "instantaneous states"). On the other hand, the simple Poisson process (Example 3 of §1.2) has this property even though it is a Markov chain on an infinite E. Ofcourse, the said property is weaker than the "almost sure continuity" of sample functions, which means that almost every sample function is a continuous function. The latter is a rather special situation in the theory of Markov

68

2. Basic Properties

processes, but is exemplified by the Brownian motion (Example 4 of §1.2) and will be discussed in §3.1. The next result is the left-handed companion to Theorem 1 of §2.3. The reader should observe the careful handling of the values 0 and 00 for the T below-often a nuisance which cannot be shrugged off.

Theorem 3. For each predictable T, we have for each fEe, and u :?: 0: (5)

where X o - = X o. Proof. Remember that lim t --+ CXJ X t need not exist at all, so that X T- is undefined on {T= oo}. Since TE:#'T_, we may erase the two appearances of l{T
Since Tn E :#'Tn , we may multiply both members of (6) by l{Tn "M}, where M is a positive number, to obtain (7)

Since Tn = 0 on {T = O}, Tn · f( 11m XT n--+CXJ

Ii Ton {T > a}, andfis continuous, we have )_{f(X u) f(X T+ u -)

+u n

on{T=O}, on{T>O}.

Fortunately, Xu = X u- almost surely in view of Lemma 2, and so we can unify the result as f(X T + u _) on Q. On the right side of (7), we have similarly lim PJ(XTJ = PJ(X T-), n--+CXJ

because X o = X o - by convention on {T = O}, and PJ is continuous. Since also :#'Tn increases to :#'T- by Theorem 8 of §1.3, and f is bounded, we can take the limit in n simultaneously in the integrand and in the conditioning field in the left side of(7), by a martingale convergence theorem (see Theorem 9.4.8 of Course). Thus in the limit (7) becomes (8)

Letting M

~ 00

we obtain (5) with X T+u~ replacing X T+u in the left member.

69

2.4. Moderate Markov Property and Quasi Left Continuity

Next, letting u 11 0, since lim uHo X T+u- = X T by right continuity and Puf -> f by Feller property, we obtain

Now define X ro _ (as well as X ro) to be aeven where lim t --+ ro X t exists and is not equal to a. Then the factor 1{T
This is then true on {T < oo} without the arbitrary fixing of the value of X ro _. For each optional T and u > 0, T + u is predictable. Hence we have just proved that X T+u = X T + u - almost surely on {T < oo}. This remark enables us to replace X T+u- by X T+u in the left member of (8), thus completing the proof of Theorem 3. D Corollary. For Y E b:li'll,

(10) The details of the proof are left as an exercise. The similarity between (10) above and (13) of §2.3 prompts the following definition. Definition. The Markov process {Xt>'~} is said to have the moderate

Markov property iff almost all sample functions have left limits and (10) is true for each predictable time T. Thus, this is the case for a Feller process whose sample functions are right continuous and have left limits. The adjective "moderate" is used here in default of a better one; it does not connote "weaker than the strong". There is an important supplement to Theorem 3 which is very useful in applications because it does not require T to be predictable. Let {Tn} be a sequence of optional times, increasing (loosely) to T. Then T is optional but not necessarily predictable. Now for each w in Q, the convergence of Tn(w) to T(w) may happen in two different ways as described below.

Case (i). \:In: Tn < T. In this case X Tn

->

X T- if T <

00.

Case (ii). 3no: T no = T. In this case X Tn = X T for n ~ no· Heuristically speaking, T has the predictable character on the set of w for which case (i) is true. Hence the result (9) above seems to imply that in both cases we have X Tn -> X T on {T < oo}. The difficulty lies in that we do not know whether we can apply the character of predictability on a part of Q without regard to the other part. This problem is resolved in the following

70

2. Basic Properties

proof by constructing a truly predictable time, which coincides with the given optional time on the part of Q in question. Theorem 4. Let Tn be optional and increase to T. Then we have almost surely

lim X Tn

=

XT

on {T < oo}.

(11)

Proof. Put A = {\:In: Tn < T};

and define T'

n

=

{Tn 00

on {Tn < T}, on {Tn = T};

T'

= {~ on A,

on Q - A.

vv

Since {Tn < T} E $'Tn and A E $'T by Proposition 6 of §1.3, optional by Proposition 4 of §1.3. We have

A n {T < oo}

c

{T' < oo}

c

A.

T~

and T' are

(12)

Hence we have \:In:

T~

= Tn < T = T' on A.

(13)

It follows that T~ /\ n < T' and T~ /\ n increases to T' on Q. Hence T' is predictable. (Note: {T = O} c Q - A, hence T' > 0 on Q.) Applying the result (9) to T', we obtain

X T , = X T ,-

on {T' < oo};

and consequently lim X Tn = XT'- = XT'

on {T' < oo}.

(14)

n

In view of (12), (13) and (14), Equation (11) is true on A n {T < oo}. But (11) is trivially true on Q - A. Hence Theorem 4 is proved. 0

The property expressed in Theorem 4 is called the "quasi left continuity" of the process. The preceding proof shows that it is a consequence of the moderate Markov property. We are now ready to prove the optionality of the hitting time for a closed set as well as an open set. For any set A in $0 define TA(w)

= inf{t > 0 IX(t, w) E A}.

(15)

This is called the (first) hitting time of A. Compare with the first entrance time D A defined in (25) of §1.3. The difference between them can be crucial.

71

2.4. Moderate Markov Property and Quasi Left Continuity

They are related as follows: T A = lim! (s suo

+ DA

0

Os),

(16)

This follows from the observation that s

+ D A Os = 0

s

°

+ inf {t z IXes + t, w) E A}

= inf{t

z sIX(t,w) E A}.

It follows from Theorem 11 of §1.3 that if DAis optional relative to {~~}, then so is s + DAD Os for each s > 0, and consequently so is T A by Proposition 3 of §1.3. It is easy to see that the assertion remains true when {~~} is

replaced by {~:'}, for each j1. Now it is obvious that DAis the debut of the set H A = {(t,w)IX(t,w)

E

A}

(17)

which is progressively measurable relative to {~~} c {~:'}, since X is right continuous. For each t, (Q, ~i, pi') is a complete probability space as a consequence of the definition of ~i. Hence D A is optional relative to {~:'} by Theorem 3 of §1.5. However that theorem relies on another which is not given in this book. The following proof is more direct and also shows why augmentation of ~~ is needed. Theorem 5. If the Markov process {X t } is right continuous, then for each open set A, D A and TA are both optional relative to {~~}. If the process is right continuous almost surely and is also quasi left continuous, then for each closed set as well as each open set A, D A and TA are both optional relative to {~;}.

Proof. By a remark above it is sufficient to prove the results for D A- If A is open, then for each t > 0, we have the identity

{D A < t} =

U {XrEA}

(18)

YEQt

where Qt = Q n [0, t), provided all sample functions are right continuous. To see this suppose DA(W) < t, then Xes, w) E A for some s E [DA(W), t), and by right continuity X(r, w) E A for some rational r in (D A(W), t). Thusthe left member of (18) is a subset of the right. The converse is trivial and so (18) is true. Since the right member clearly belongs to ~~, we conclude that {D A < t} E ~~. Now suppose that only P-almost all sample functions are right continuous. Then the argument above shows that the two sets in (18) differ by a P-null set. Hence {D A < t} is in the augmentation of ~~ with respect to (Q, ~, P). Translating this to pi', we have {D A < t} E ~i for each j1, hence {D A < t} E ~;. Since {.?;} is right continuous we have {D A :s; t} E ~;.

72

2. Basic Properties

Next, suppose that A is closed. Then there is a sequence of open sets An such that An ::::> An + 1 for each n, and (19) n

We shall indicate this by An

n

11 A. For instance, we may take

An =

{x

E

Eold(x,A) <

~}.

Clearly DAn increases and is not greater than DA; let S = lim i DAn :s; DA-

(20)

n

We now make the important observation that for any Borel set B, we have almost surely (21) For if DB(w) < 00, then for each <5 > 0, there exists t E [DB(w), DB(w) + <5) such that X(t, w) E B. Hence (21) follows by right continuity. Thus we have X(D AJ EAn for all n and therefore by quasi left continuity and (19): X(S) = lim X(DAJ n

En An = A n

almost surely on {O:S; S < oo}. The case S = 0 is of course trivial. This implies S ~ D A; together with (20) we conclude that D A = S a.s. on {S < oo}. But D A ~ S = 00 on {S = oo}. Hence we have proved a.s. (22) Since each DAn is optional relative to {ff;}; so is DA by (22).

o

If we replace the D's by T's above, the result (22) is no longer true in general! The reader should find out why. However, since for each Borel set B, DB = TB on {X o ¢ B}, we have the following consequence of (22) which is so important that we state it explicitly and more generally as follows. Corollary. Let An be closed sets, An::::> A n+ 1 for all n, and X

¢ A, then PX-a.s. we have

TA = lim n

i TAn:S; + 00.

Here A may be empty, in which case TA =

+ 00.

nn An = A. If (23)

73

2.4. Moderate Markov Property and Quasi Left Continuity

Exercises

1. Here is a shorter proof of the quasi left continuity of a Feller process (Theorem 4). For rt. > 0, IE bC+, {e-"tU"/(X t)} is a right continuous positive supermartingale because U"f is continuous. Hence if Tn i T, and Y = limn X(T n) (which exists on {T < oo}), we have a.s. U"I(Y)

=

li~ U"f(X(Tn)) = E {U"/(X(T)) Iy g;Tn }.

Now multiply by

rt.,

let rt. --->

00,

and use Lemma 1.

2. Prove the following "separation" property for a Feller process. Let K be compact, G be open such that KeG. Then for any 0 > 0 there exists to > 0 such that inf r{TGc ~ to} ~ 1 -

o.

XEK

[Hint: let 0 :::;; I:::;; 1, 1= 1 on K, 1=0 in GC. There exists to> 0 such that supo:O:t:O:to Ilpt! - III < 0/2. Now consider

and apply Theorem 3 of §2.3.] 3. If Y E g;-/fff' and IE fff'-/PlJ then I(Y) E g;-/PJJ. Hence x ---> EX{f(Y)} is in fff'- if the expectation exists. [This will be needed for Lebesgue measurability for important functions associated with the Brownian motion.] To appreciate the preceding result it is worthwhile to mention that a Lebesgue measurable function of a Borel measurable function need not be Lebesgue measurable (see Exercise 15 in §1.3 of Course). Thus if we take g;O = PlJ and Y E g;O; if I E g;/l we cannot infer that I(Y) E g;/l, where fl is the Lebesgue measure on g;o.

NOTES ON CHAPTER

2

§2.1. This chapter serves as an interregnum between the more concrete Feller processes and Hunt's axiomatic theory. It is advantageous to introduce some of the basic tools at an early stage. §2.2. Feller process is named after William Feller who wrote a series of pioneering papers in the 1950's. His approach is essentially analytic and now rarely cited. The sample function properties of his processes were proved by Kinney, Dynkin, Ray, Knight, among others. Dynkin [1] developed Feller's theory by probabilistic methods. His book is rich in content but difficult to consult owing to excessive codification. Hunt [1] and Meyer [2] both discuss Feller processes before generalizations. §2.3. It may be difficult for the novice to appreciate the fact that twenty five years ago a formal proof of the strong Markov property was a major event. Who is now interested in an example in which it does not hold?

74

2. Basic Properties

A full discussion of augmentation is given in Blumenthal and Getoor [1]. This is dry and semi-trivial stuff but inevitable for a rigorous treatment of the fundamental concepts. Instead ofbeginning the book by these questions it seems advisable to postpone them until their relevance becomes more apparent. §2.4. There is some novelty in introducing the moderate Markov property before quasi left continuity; see Chung [5]. It serves as an illustration of the general methodology alluded to in §1.3, where both F T+ and F T- are considered. Historically, a moderate Markov property was first observed at the "first infinity" of a simple kind of Markov chains, see Chung [3]. Indeed, it occurred as a sudden revelation at the last stage of page-proofing the manuscript. It turns out that a strong Markov process becomes moderate when the paths are reversed in time, see Chung and Walsh [I]. A more complete discussion of the measurability of hitting times will be given in §3.3. Hunt practically began his great memoir [1] with this question, fully realizing that the use of hitting times is the principal method in Markov processes.

Chapter 3

Hunt Process

3.1. Defining Properties Let {Xt, g;, t E T} be a (homogeneous) Markov process with state space (Eo, 0"0) and transition function (Pt), as specified in §1.1 and §1.2. Here g; is the fF; defined in §2.3. Such a process is called a Hunt process iff (i) (ii) (iii)

it is right continuous; it has the strong Markov property (embodied in Theorem 1 of §2.3); it is quasi left continuous (as described in Theorem 4 of §2.4).

Among the basic consequences ofthese hypotheses are the following: (iv) (v)

{g;} is right continuous (Corollary to Theorem 4 of §2.3); {X t } is progressively measurable relative to {g;} (Theorem 1 of

(vi)

(P t ) is Bore1ian (Exercise 4 of §1.5).

§1.5); We have shown in Chapter 2 that given a Feller semigroup (Pt), a Feller process can be constructed having all the preceding properties (and others not implied by the conditions above). Roundly stated: a Feller process is a Hunt process. Whereas a Feller process is constructed from a specific kind of transition function, a Hunt process is prescribed by certain hypotheses regarding the behavior of its sample functions. Thus in the study of a Hunt process we are pursing a deductive development of several fundamental features of a Feller process. To begin with, we can add the following result to the above list of properties of a Hunt process. Theorem 1. Almost surely the sample paths have left limits in (0, CD). Proof. For a fixed

I::

> 0, define (1)

where d denotes a metric of the space Eo. Our first task is to show that T is optional, indeed relative to {gpn For this purpose let {Zk} be a countable

76

3. Hunt Process

dense set in Ea, and B kn be the closed ball with center Zk and radius n -1 :Bkn = {xld(x,Zk)::; n- 1}. Then {B kn } forms a countable base of the topology. Put

Then T kn is optional relative to {ff~} by Theorem 5 of §2.4, since the set {x Id(x, B kn ) > e} is open. We claim that {T < t} =

U{X o E B kn ; T kn < t}.

(2)

k.n

It is clear that right member of (2) is a subset of the left member. To see the converse, suppose T(w) < t. Then there is 8(W) < t such that d(Xs(w), Xo(w)) > e; hence there are nand k (both depending on w) such that d(Xs(w), Xo(w)) > e + 2n- 1 and Xo(w) E Bkn' Thus d(Xs(w), B kn ) > e and Tkn(w) < t; namely w belongs to the right member of (2), establishing the identity. Since the set in the right member of (2) belongs to ff~, T is optional as claimed. Next, we define To == 0, T 1 == T and inductively for n ;::.: 1:

Each Tn is optional relative to {ff?} by Theorem 11 of §1.3. Since Tn increases with n, the limit S = limn Tn exists and S is optional. On the set {S < oo}, we have limnX(Tn) = X(S) by quasi left continuity. On the other hand, right continuity of paths implies that d(X(Tn+d, X(Tn)) ;::.: e almost surely for all n, which precludes the existence oflimnX(Tn). There would be a contradiction unless S = 00 almost surely. In the latter event, we have [0,00) = U:,=o [Tn, T n+1)' Note that if Tn = 00 then [T m T n+1) = 0. In each interval [Tn, T n+1) the oscillation of XC) does not exceed 2e by the definition of T n + l' We have therefore proved that for each e, there exists QE with P(QJ = 1 such that X( .) does not oscillate by more than 2e in [T~, T~+ d, where ro

[0,00)=

U [T~,T~+1)'

(3)

n=O

n:=

Let Q. = 1 Q1/m; then P(Q.) = 1. We assert that if WE Q., then XL w) must have left limits in (0,00). For otherwise there exists t E (0, 00) and m such that X(', w) has oscillation > 21m in (t - (j, t) for every (j > 0. Thus t ¢ [T~/m, T~~m1) for all n ;::.: 0, which is impossible by (3) with e = 11m. 0

n

Remark. We prove later in §3.3 that on {t < we have X t - #- 0, namely X t - E E. For a Feller process this is implied by Theorem 7 of §2.2. Let us observe that quasi left continuity implies that X( .) is left continuous at each fixed t E (0, 00), almost surely. For if the tn's are constant such that t n if t, then each t n is optional and so X(tn) -4 X(t). Coupled with right continuity, this implies the continuity of almost all paths at each fixed t. In

°::;

77

3.1. Defining Properties

other words, the process has no fixed time of discontinuity. For a Feller process, this was remarked in §2.4. Much stronger conditions are needed to ensure that almost all paths are continuous. One such condition is given below which is particularly adapted to a Hunt process. Another is given in Exercise 1 below. Theorem 2. Let {X t} be a Markov process with right continuous paths having left limits in (0, 00). Suppose that the transition function satisfies the following condition: for each e > and each compact K c E we have

°

lim ~ sup [1 - Pt(x, B(x, e))] = t~O t XEK where B(x, e)

= {y E Eald(x, y):::;; e}.

°

(4)

Then almost all paths are continuous.

o

Proof. The proof depends on the following elementary lemma.

Lemma. Let f be a function from [0,1] to Ea which is right continuous in [0, 1) and has left limits in (0, 1]. Then f is not continuous in [0,1] (continuity at the endpoints being defined unilaterally) if and only if there exists e > such that for all n ;::: no(e) we have

°

max d (f (~),f (~)) > e. O$k$n-1 n n

(5)

Proof of the Lemma. If f is not continuous in [0, 1], then there exists t E such that d(f(t - ), f(t)) > 2e. For each n ;::: 1 define k by kn- 1 < t :::;; (k + 1)n- 1. Then for n ;::: no(e) we have d(f(kn -1),f(t-)) < e/2, d(f( (k + 1)n- 1),f(t)) < e/2; hence d(f(kn -1 ),f( (k + l)n -1)) > e as asserted.

(0,1] and e >

°

Conversely, if f is continuous in [0, 1], then f is uniformly continuous there and so for each e > (5) is false for all sufficiently large n. This is a stronger conclusion than necessary for the occasion. To prove the theorem, we put for a fixed compact K:

°

M=

M: =

{wIX(·,w) is not continuous in [0,1]; X(s,w)

E

{wIO$~~~_1 d( xG'w)' x(k: 1, w)) >e}; X(s,w)

E

K for all s E [0,1]}-

It follows from the lemma that M

E ffo,

and

K for all s E [0, 1]},

78

3. Hunt Process

If we apply the Markov property at all kn -1 for 0 :s:: k :s:: n - 1, we obtain

:s:: n sup P l/nCX, B(x, xEK

2n

U sing the condition (4) with t = n- 1, we see that the last quantity above tends to zero as n -400. Hence P(liminfn M~):S:: limn P(M~) = 0 and P(M) = 0. Now replace the interval [0,1] by [1/2,1/2 + 1] for integer I 2 1, and K by K m U {a} where K m is a compact subset of E and K m l' E. Denote the resulting M by M(I, m) and observe that we may replace K by K {a} in (4) because ais an absorbing state. Thus we obtain

U

which is seen to be equivalent to the assertion of the theorem.

o

As we have seen in §2.5, the Brownian motion in R 1 is a Feller process. We have

EXAMPLE.

Pt(x, B(X,2n

=

1 ~

y2nt

~ Iy-xl>e

exp [ - (y 2t

[u

xfJ

dy

2

2 Jeroo exp - 2t J du = J2ij

= '.)f2 Jeroo ~t ( exp

m

2J [u J u) f2 t [2 - 2t t du:S:: '.) mEexp - 2t . 2

Hence (4) is satisfied even if we replace the K there by E = R 1 . It follows that almost all paths are continuous. Recall that we are using the fact that the sample paths of a [version of] Feller process are right continuous in [0,(0) and has left limits in (0,00), which was proved in §2.2. Thus the preceding proofis not quite as short as it appears. The result was first proved by Wiener in 1923 in a totally different setting. Indeed it pre-dated the founding of the theory of stochastic processes. Exercises

1. A stochastic process {X(t), t 2 O} is said to be separable iff there exists a countable dense set S in [0, (0) such that for almost every w, the sample function X(·, w) has the following property. For each t 2 0, there exists Sn E S such that Sn -4 t and X(sm w) -4 X(t, w). [The sequence {sn} depends

79

3.1. Defining Properties

on w as well as t!] It is an old theorem of Doob's that every process has a version which is separable; see Doob [1]' If {X(t)} is stochastically continuous the proof is easy and the set S may be taken to be any countable dense set in [0,00). Now take a sample function XC, w) which has the separability property above. Show that if X(s, w) with s restricted S is uniformly continuous in S, then X(t, w) is continuous for all t ;:::: O. Next, suppose that there exist strictly positive numbers 15, IX, f3 and C such that for t ;:::: 0 and 0 < h < 15 the following condition is satisfied:

Take S to be the dyadics and prove that X(s, w) with s E S is continuous on S for almost every w. Finally, verify the condition above for the Brownian motion in R 1. [Hint: for the last assertion estimate 2:l:(/ P{IX((k + 1)r n) - X(k2~n)1 > n- 2 } and use the Borel-Cantelli lemma. For two dyadics sand s', X(s) - X(s') is a finite sum of terms of the form X((k + 1)r n) - X(kr n). The stated criterion for continuity is due to Kolmogorov.] 2. Give an example to show that the Lemma in this section becomes false if the condition "f has left limits in (0, 1]" is dropped. This does not seem easy, see M. Steele [1]. 3. Let X be a homogeneous Markov process. A point x in E is called a "holding point" iff for some 15 > 0 we have Px {X(t) = x for all t E [O,15]} > O. Prove that in this case there exists A. ;:::: 0 such that PX{T{xjC > t} = e- M for all t > o. When A. = 0, x is called an "absorbing point". Prove that if X has the strong Markov property and continuous sample functions, then each holding point must be absorbing. 4. For a homogeneous Markov chain (Example 1 of §1.2), the state i is holding (also called stable) if and only if

. 1 - Pii(t) 11m < 00. tto

t

The limit above always exists but may be + 00, when finite it is equal to the A. in Problem 3. In general, a Markov chain does not have a version which is right continuous (even if all states are stable), hence it is not a Hunt process. But we may suppose that it is separable as in Exercise 1 above. 5. For a Hunt process: for each x there exists a countable collection of optional times {Tn} such that for PX-a.e. w, the set of discontinuities Tn(w). [Hint: for each e> 0 define s(e) = of X(·, w) is the union inf{t > O!d(Xt_,X t) > e}. Show that each s(e) is optional. Let s~) = s(e), s~eL = s~e) + s(e) 0 e(s~e)) for n ;:::: 1. The collection {s~l/m)}, m ;:::: 1, n ;:::: 1, is the desired one.]

Un

80

3. Hunt Process

6. Let T be one of the Tn's in Exercise 5. Suppose that R n is an increasing sequence of optional times such that r{lim n R n = T < oo} = 1, then r{Un(R n = T)} = 1. This means intuitively that T cannot be predicted. A stronger property (called "total inaccessibility") holds to the effect that for each x the two probabilities above are equal for any increasing sequence of optional {R n} such that Rn:S; T for all n; see Dellacherie and Meyer [1]' 7. Prove that for a Hunt process, we have !FT = O"(!FT-,X T). (See Chung

[5].)

3.2. Analysis of Excessive Functions A basic tool in the study of Hunt processes is a class of functions called by Hunt "excessive". This is a far-reaching extension of the class of superharmonic functions in classical potential theory. In this section simple basic properties of these functions will be studied which depend only on the semigroup and not on the associated Markov process. Deeper properties derived from the process will be given in the following sections. Let a Borelian transition semigroup (P t , t :2: 0) be given where Po is the identity. For a :2: 0 we write (1)

thus P7 = Pt. For each a, (P~) is also a Borelian semigroup. It is necessarily submarkovian for a > O. The corresponding potential kernel is U~, where (2)

Here f E bl!' + or f E I!' +. If f E bl!'+ and a > 0, then the function UJ is finite everywhere, whereas this need not be true if f E I!' +. This is one reason why we have to deal with U~ sometimes even when we are interested in UO = U. But the analogy is so nearly complete that we can save a lot of writing by treating the case a = 0 and claiming the result for a :2: O. Only the finiteness of the involved quantities should be watched, and the single forbidden operation is "00 - 00". The a-potential has been introduced in §2.1, as are a-superaveraging and ex-excessive functions. The class of a-excessive functions will be denoted by S~, and SO = S. We begin with a fundamental lemma from elementary analysis, of which the proof is left to the reader as an essential exercise. Lemma 1. Let {u mn } be a double array of positive numbers, where m and n are positive integer indices. If U mn increases with n for each fixed m, and increases

81

3.2. Analysis of Excessive Functions

with m for each fixed n, then we have

lim lim U mn = lim lim U mn = lim u nn . m

n

n

Of course these limits may be

m

n

00.

Proposition 2. For each rx 2': 0, S~ is a cone which is closed under increasing sequential limits. The class of rx-superaveraging functions is also such a cone which is furthermore closed under the minimum operation" /\". Proof. It is sufficient to treat the case rx = O. To say that S is a cone means: if /; E S, and Ci are positive constants for i = 1, 2, then cd! + c 2f2 E S. This is trivial. Next let fn E Sand fn i f. Then for each nand t

letting n ~ 00 we obtain f 2': PJ by monotone convergence. If f is superaveraging, then PJ increases as t decreases. Hence if fn E Sand fn i f, we have f = lim fn = lim lim PJn = lim lim PJn = lim PJ, n

t n t

t

by Lemma 1 applied to n and a sequence of t (> 0) decreasing to O. This proves f E S. Finally, if /; is superaveraging for i = 1,2, then /; 2': Pt/; 2': Pt(ft /\f2); hence ft /\f2 2': Plft /\f2); namely that ft /\f2 is superaveraging. 0 It is remarkable that no simple argument exists to show that S is closed under" /\ ". A deep proof will be given in §3.4. The following special situation is very useful. Proposition 3. Suppose P t converges vaguely to Po as t ! O. If f is superaveraging and also lower semi-continuous, then f is excessive. Proof. By a standard result on vague convergence, we have under the stated hypotheses: f

= Pof ~ lim PJ ~ f. tto

o

If (Pt) is Fellerian, then the condition on (Pt) in Proposition 3 is satisfied. More generally, it is satisfied if the sample paths of the associated Markov process are right continuous at t = O. For then if f E be, as t! 0:

Here we have a glimpse of the interplay between the semigroup and the process. Incidentally, speaking logically, we should have said "an associated Markov process" in the above.

82

3. Hunt Process

Proposition 4. If

IX

< [3, then sa

c SP.

For each

IX ;;:::

0, sa

=

nP>a Sp.

Proof· It is trivial that if IX < [3, then P~f;;::: Nffor f E tff+. Now if f = limtto PUfor any value of [3 ;;::: 0, then the same limit relation holds for all values of [3 ;;::: 0, simply because lim tto e- Pt = 1 for all [3 ;;::: 0. The proposition follows quickly from these remarks. 0

The next result gives a basic connection between superaveraging functions and excessive functions. It is sufficient to treat the case IX = 0. If f is superaveraging, we define its regularization] as follows: ](x) = lim ttO

i PJ(x).

(4)

We have already observed that the limit above is monotone. Hence we may define Pt+f(x) = limsHt Psf. Proposition 5. ]

E

Sand ] is the largest excessive function not exceeding

f.

We have (5)

Proof. We prove (5) first, as follows:

It is immediate from (4) that]:::; f and];;::: PrJ;;::: PrJ. Furthermore by (5),

lim tHO

prJ = lim Pt+f = lim PrJ = ], tHO

tHO

where the second equation is by elementary analysis; hence] E S. If g E S and g :::; f, then Ptg :::; PrJ; letting t 11 we obtain g :::;]. 0

°

The next result contains an essential calculation pertaining to a potential. Theorem 6. Suppose f

E

S, PrJ <

00

for each t > 0, and

lim PrJ = 0.

(6)

t-+ 00

Then we have (7)

83

3.2. Analysis of Excessive Functions

Proof. The hypothesis Ptf <

00

allows us to subtract below. We have for

h > 0:

Jor Ps(f t

Phf)ds =

Jor

r

t

Jor PJ ds -

t+ h PJ ds - Jh PJ ds =

i

If we divide through by h above, then let t zero by (6) and we obtain

00,

h

rt

Jt

+h

PJ ds.

the last term converges to

(8)

The integrand on the left being positive because f ~ Pd, this shows that the limit above is the potential shown in the right member of (7). When h 1 0, the right member increases to the limit f, because limsto i PJ = f· This establishes (7). 0 Let us observe that there is an obvious analogue of Proposition 6 for and f E blff +, then the corresponding conditions in the proposition are satisfied. Hence (7) holds when U and Ph are replaced by ua and Ph for such an f. This is an important case of the theorem. An alternative approach to excessive functions is through the use of resolvents (U a ) instead of the semigroup (P t ). This yields a somewhat more general theory but less legible formulas, which will now be discussed briefly. We begin with the celebrated resolvent equation (9) below.

sa. If rt. >

°

Proposition 7. For

rt.

>

°

and f3 > 0, we have (9)

Proof. This is proved by the following calculations, for f UaUPf =

E

blff+ and

Sooo e-asp s[Sooo e-PtPtf dtJ ds

Joroo Joroo e-as-ptps+t fdtds = Sooo 1 e
=

00

=

=

s + t,

Sooo [S: e(p-alsds]e-puPufdU au

=

where u

roo e- Jo f3 -

ert.

PU

Puf du

=

~1_ [uaf - UPf]. f3 - rt.

rt. "'"

f3:

84

3. Hunt Process

Note that the steps above are so organized that it is immaterial where (J - rt. > 0 or < O. This remark establishes the second as well as the first equation in (9). 0 If f

E

S, then since

we have (10) lim

l' rt.Ul1.f = f·

(11)

11.; 00

[The case f(x) = tricky to prove.

00

in (11) should be scrutinized.] The converse is rather

PropositionS. If fErff+ and (10) is true, thenf*=liml1.;oo 1'rt.Ul1.fexistsand is the largest excessive function ~f If (11) is also true then f is excessive. Proof. Let 0 <

p<

rt.

and f

E

brff +. Then we have by (9): (12)

Here subtraction is allowed because Uaf < 00. If (10) is true, then gap = f - (rt. - (J)U,,! ;;::: O. Since UP gl1.P is (J-excessive, Uaf is (J-excessive for all (J E (0, rt.). Hence U"! E S by Proposition 4. Next, we see from (12) by simple arithmetic that

by (10). Hence if we put

f* = lim 1'aU"! = lim l' UI1.(rt.f), 11.;00

11.;00

we have f* E S by Proposition 2. For a general f E rff + satisfying (10), let fn = f /\ n. Then rt.U"!n :::;; fn by (10) and the inequality rt.Ul1.n :::;; n. Hence we may apply what has just been proved to each fn to obtain f: = lima; 00 1'rt.Ul1.fn. Also the inequality rt.Ul1.fn ;;::: (JUPfn leads to rt.U"! ;;::: PUPf, if rt. ;;::: (J. Hence we have by Lemma 1: f* = lim

l' rt.Ul1.f = lim 1'lim l' rt.Ul1.fn

11.;00

=

lim 1'lim n;oo 11.;00

11.;00

n;oo

l' rt.Ul1.fn = lim l' n. n;oo

85

3.2. Analysis of Excessive Functions

Since each

f~ E

S as shown above, we conclude that f* E S. If 9 E Sand i 00 we obtain 9 ::; f*· This identifies f* as the largest excessive function::; f. The second assertion of Proposition 8 is now trivial.* 0

9 ::; f, then aUag ::; aUaf; letting a

f

Proposition 8 has an immediate extension as follows. For each f3 SfJ if and only if

~

0,

E

Va> 0: aUa+fJf ::; f

and

lim aj

i

aUa+fJf = f·

(14)

00

This is left as an exercise. The next result is extremely useful. It reduces many problems concerning an excessive function to the same ones for potentials, which are often quite easy. Theorem 9. If f

E

S, then for each f3 > 0, there exists gn

E

bi!+ such that

(15) n

Proof. Suppose first that f

E bi! +. Then as in the preceding proof, we have Ua(af) = UfJ(agafJ)' Thus we obtain (15) from (11) with gn = ngnD' In general we apply this to obtain for each k, = limn i UfJg~k). It follows then by Lemma 1 that

n

f = f* = lim

n = lim i lim i

k

k

UfJg~k)

n

= lim i

UfJg~n).

0

n

We remark that this proposition is also a consequence of the extension of Theorem 6 to f3-excessive functions, because then lim t ... 00 Nf = 0 for f E hi! +, etc. But the preceding proof is more general in the sense that it makes use only of (10) and (11), and the resolvent equation, without going back to the semlgroup. It is an important observation that Theorem 9 is false with f3 = 0 in (15); see Exercise 3 below. An additional assumption, which turns out to be of particular interest for Hunt processes, will now be introduced to permit f3 = 0 there. Since we are proceeding in an analytic setting, we must begin by assuming that Vx

E

E: U(x,E) > 0.

(16)

This is trivially true for a process with right continuous paths; for a more general situation see Exercise 2 below. Remember however that U(8, E) = O. To avoid such exceptions we shall agree in what follows that a "point" means a point in E, and a "set" means a subset of E, without specific mention to the *1 am indebted to Patrick Fitzsimmons for correcting a slip in the first edition.

86

3. Hunt Process

contrary. The case of 8 can always be decided by an extra inspection, when it is worth the pain. Definition. The semigroup (Pt) or the corresponding U is called transient iff there exists a function h E iff + such that

0< Uh <

(17)

on E.

00

An immediate consequence of (17) is as follows. There exists a sequence of hn E M + (with hn(8) = 0) such that Uh n > 0

Uh n i

and

00

both in E.

(18)

We may take hn = nh to achieve this. The probabilistic meaning of transience as defined above, as well as some alternative conditions, will be discussed later in §3.7. Proposition 10: If (P t ) is transient, and f that

f =

lim

i

E

S, then there exists gn

Ugn-

E

biff + such (19)

n

In fact, gn :s; n 2 and Ug n :s; n.

Proof. Put fn = f /\ Uh n /\ n where h n is as in (18). Then we have

as t i 00 because Uh n < 00. By Proposition 2, fn is superaveraging; let its regularization be In. Since Ptln :s; Ptf~ < 00, and --+ 0 as t i 00, we may apply Theorem 6 to obtain ln = lim

i

Ug nk

k

where

From the proof of Theorem 6 we see that

87

3.3. Hitting Times

For each n, Ug nk increases with k; for each k, Ug nk increases with n, hence we have by Lemma 1: lim

i

1:. =

lim

i

lim

n

i

k

Ug nk

=

lim n

i

Ug nw

On the other hand, we have again by Lemma 1: lim n

i

1:. = lim i n

lim tto

i

Ptfn = lim tto

i

lim n

i

Ptfn = lim Ptf =

no

f·

o

It follows that (19) is true with gn = gnw

Exercises 1. A measure J.l on cS is said to be c-finite ("countably finite") iff there is a 1 J.ln(A) countable collection of finite measures {J.ln} such that J.l(A) = for all A E cS. Show that if f.l is IT-finite, then it is c-finite, but not vice versa. For instance, Fubini's theorem holds for c-finite measures. Show that for each x, U(x,·) is c-finite.

L:7=

2. Show that for each x E E, Pt(x, E) is a decreasing function of t. The following properties are equivalent: (a) U(x,E) = 0;

(b) lim Pt(x, E) = 0;

(c)

rg = O} = 1.

tHO

3. For the Brownian motion in R 1 , show that there cannot exist positive Borel (or Lebesgue) measurable functions gn such that limn~w Ug n = 1. Thus Theorem 9 is false for f3 = O. [See Theorem 1 of §3.7 below for general elucidation.J 4.

Iff E cS + and aU,,! :s;; f for all a > 0, then Ptf :s;; f for (Lebesgue) almost all t

~

O. [Hint: limajw aUaf = f* :s;; f and U"!* = uaf for all a > O.J

3.3. Hitting Times We have proved the optionality of T A for an open or a closed set A, in Theorem 5 of§2.4. The proof is valid for a Hunt process. It is a major advance to extend this to an arbitrary Borel set and to establish the powerful approximation theorems "from inside" and "from outside". The key is the theory of capacity due to Choquet. We begin by identifying the probability of hitting a set before a fixed time as a Choquet capacity. Let {XI> ff';, t E T} be a Hunt process as specified in §3.1. For simplicity of notation, we shall write P for pI-' (for an arbitrary probability measure J.l on cS); but recall that ff'; c :Fr and (:FI-', pI-') is complete. Fix t ~ O. For any subset

88

3. Hunt Process

A of Eo, we put

A' = {w

E

QI:ls

E

[O,t]:Xs(W)

E

A}.

This is the projection on Q of the set of (s, w) in [0, t] x Q such that X(s, w) E A. The mapping A --+ A' from the class of all subsets of Eo to the class of all subsets of Q has the following properties: (i) (ii) (iii)

A 1 C A 2 => A'l C A 2; (A 1 U A 2 )' = A'l U A 2; An i A => A~ i A'.

In view of (i) and (ii), (iii) is equivalent to

(iii')

(Un An)' = Un A~ for an arbitrary sequence {An}·

So far these properties are trivial and remain valid when the fundamental interval [0, t] is replaced by [0, CX)). The next proposition will depend essentially on the compactness of [0, t]. We shall write A 1 c:: A 2 to denote P(A 1\A 2 ) = 0, and A 1 == A 2 to denote P(A 1 6. A 2) = 0. If G is open, then G' E:F°, so peG') is defined. Proposition 1. Let Gn be open, K be compact such that (1)

Then we have K'

==

nG~.

(2)

n

Proof. This is essentially a re-play of Theorem 5 of §2.4. Consider the first entrance times DGn ; these are optional (relative to {:F?}). If WE nn G~, then Ddw) :s; t for all n. Let D(w) = limn i Ddw); clearly D(w) :s; DK(w) and D(w) :s; t. Put

°

we have P(N) = by quasi left continuity. If WE nn G~\N, then X(D) E nn Gn = K since X(DGJ E Gn by right continuity. Thus DK(w) :s; D(w) and so DK(w) = D(w) :s; t; namely WE K'. We have therefore proved nn G~ c:: K'. But nn G~ ::J K' by (i) above, hence (2) is proved. D Proposition 2. Let K n be compact, K n ! and nn K n K'

==

n

K~.

n

= K. Then (3)

89

3.3. Hitting Times

Proof. K being given, let {G n} be as in Proposition 1. For each m, G m =:J K = K n ; hence by a basic property of compactness, there exists nm such that G m =:J K nm and consequently G;" =:J K~m by (i). It follows that

nn

K'

==

n G;" n K~m n K~ =:J

m

=:J

=:J

K',

m

o

where the first equation is just (2). This implies (3).

Corollary. For each compact K, and 8> 0, there exists open G such that KeG and P(K')

s

P(G')

s

P(K')

+ 8.

In particular P(K')

= inf P(G')

(4)

G=>K

where the inf is over all open G containing K.

From here on in this section, the letters A, B, G, K are reserved for arbitrary, Borel, open, compact sets respectively. Definition. Define a set function C for all subsets of E a as follows: (a) (b)

C(G) = P(G'); C(A) = infG=>A C(G).

Clearly C is "monotone", namely A l c A z implies C(AJ that it follows from (4) that C(K)

s

C(A z ). Note also

= P(K').

(5)

We proceed to derive properties of this function C. Part (b) of its definition is reminiscent of an outer measure which is indeed a particular case; on the other hand, it should be clear from part (a) that C is not additive on disjoint sets. What replaces additivity is a strong kind of subadditivity, as given below. Proposition 3. C is "strongly subadditive" over open sets, namely; (6) Proof. We have, as a superb example of the facility (felicity) of reasoning with sample functions:

(G l

U

Gz)' - G'l

=

{wl3s E [O,t]:Xs(w) E G l u Gz ; '<:Is E [O,t]:Xs(w) ¢: Gd c {w 13s E [0, t]:Xs(w) E G z ; '<:Is E [0, t]: Xs(w) ¢: G l n Gz}

= G~

- (G 1 n G z )'.

90

3. Hunt Process

Taking probabilities, we obtain

Up to here GI and Gz may be arbitrary sets. Now we use definition (a) above to convert the preceding inequality into (6). 0 The same argument shows C is also strongly additive over all compact sets, because of (5). Later we shall see that C is strongly additive over all capacitable sets by the same token.

Lemma 4. Let AnI, An

C

Gn, en > 0 such that

(7) Then we have for each finite m:

(8)

Proof. For m = 1, (8) reduces to (7). Assume (8) is true as shown and observe

that (9)

We now apply the strong subadditivity of C over the two open sets and Gm + I' its monotonicity, and (9) to obtain

CCv:

Gn)

+ C(A m) ~

C((9I

~ C(DI Gn) + C(G

m+ l )

Gn)

U

Gm + l )

+

C((9I

~ C(A m) + JI en + C(Am+

U::,=

l)

+ em+ l ·

We can now summarize the properties of C in the theorem below.

(i) Al c A z => C(A I ) ~ C(A z ); (ii) An i A=> C(A n) i C(A); (iii) K n t K => C(K n ) t C(K).

Gn

Gn) n Gm + l )

This completes the induction on m.

Theorem 5. We have

I

o

91

3.3. Hitting Times

Proof. We have already mentioned (i) above. Next, using the notation of Lemma 4, we have

by definition (a) of C, and the monotone property of the probability measure P. Applying Lemma 4 with 8 n = 82- n, we see from (8) that m

Since 8 is arbitrary, (ii) follows from this inequality and (i). Finally, it follows from (5), Proposition 2, and the monotone property of P that C(K) = P(K') = P

(n K~) =

lim P(K;,) = lim C(Kn).

n

n

n

This proves (iii).

D

Definition. A function defined on all the subsets of Eo and taking values in [ - 00, + 00] is called a Choquet capacity iff it has the three properties (i), (ii) and (iii) in Theorem 5. A subset A of Eo is called capacitable iff given any 8

> 0, there exist an open G and a compact K such that K cAe G and C(G) ::; C(K)

+

8.

(10)

This definition (which can be further generalized) is more general than we need here, since the C defined above takes values in [0,1] only. Here is the principal theorem of capacitability. Choquet's Theorem. In a locally compact Hausdorff space with countable base, each analytic set is capacitable.

A set A c Eo is analytic iff there exists a separable complete metric space (alias "Polish space") M and a continuous mapping

C(B)

=

P(B').

(11)

92

3. Hunt Process

Proof. Since B is capacitable by Choquet's theorem, for each n ~ 1 there exist K n c B c Gn such that C(K n ) S C(B) S C(G n) S C(Kn)

We have

K~ c

G~.

B' c

UK~ c

B' c

n

~

(12)

n

Let

Al =

Then for each n

+~.

n G~ = A z·

(13)

n

1:

and consequently (14)

It follows that P(B')

=

P(A I )

= lim P(K~) = lim C(K n) = n

C(B)

n

o

where the last equation is by (12). Corollary. P(B') = lim

= lim P(G~).

P(K~)

n

n

The optionality of DB and T B follows quickly. Theorem 7. For each B Ega,

DB

and T B are both optional relative to {~}.

Proof. Recall that t is fixed in the foregoing discussion. Let us now denote the B' above by B'(t). For the sake of explicitness we will replace (Q,:!F, P) above by (Q,~, PJi) for each probability measure J.l on g, as fully discussed in §2.3. Put also Ql = (Q n [O,t)) u {t}. Since G is open, the right continuity of s ~ X s implies that G'(t)

= {wl3s E QI:Xs(W) E G}.

°

nn

Hence G'(t) E:!F~ c :!Fi. Next, we have pJi{B'(t) L,. G~(t)} = by (13), hence B'(t) E:!Fi because :!Fi contains all PJi-null sets. Finally, a careful scrutiny shows that for each B E ga and t > 0: {wIDB(w) < t} =

U U

{wl3s

E

[O,r]:Xs(w)

rEQn[O,I)

rEQn[O,t)

B'(r)

E

:!Fi·

E

B}

93

3.3. Hitting Times

Hence DB is optional relative to of §2.4.

{~i}.

The same is true of T B by using (16) D

It should be recalled that the preceding theorem is also an immediate consequence of Theorem 3 of §1.5. The more laborious proof just given "tells more" about the approximation of a Borel set by compact subsets and open supersets, but the key to the argument which is concealed in Choquet's theorem, is the projection from T x Q to Q, just as in the case of Theorem 3 of §1.5. This is also where analytic sets enter the picture; see Dellacherie and Meyer [1]' The approximation theorem will be given first for DB' then for T B, in two parts. The probability measure /1 is defined on rffo, though the point aplays only a nuisance role for what follows and may be ignored.

Theorem 8(a). For each /1 and BE rffo, there exist K n c B, K ni and G n :::> B, Gnl such that

(15)

P"-a.s.

Proof. The basic idea is to apply the Corollary to Theorem 6 to B'(r) for all

r E Q. Thus for each r E Q, we have (16)

such that lim P"{ G~ir) -

K~n(r)} =

(17)

O.

n

Let {rJ be an enumeration of Q, and put n

Kn =

UK

n

rjn ,

Gn =

j= 1

n Grjn'

j= 1

Then D' 2 DB 2 D". We have for eachj: {D' > rj > D"} c

n{D

Kn

> rj> Dc,,}

c

n{G~(r) n

For n 2 j, we have hence

K~(r)}.

(18)

94

3. Hunt Process

Hence by (17), lim pl'{ G~(r) -

K~(r)} =

0

n

and consequently by (18), PI'{D' > rj > D"} = O. This being true for all rj , we conclude that pl'{D' = DB = D"} = 1, which is the assertion in (15). 0 Theorem 8(b). For each p. and BE go, there exists K n c B,Knj such that

(19) If P. is such that p.(B) = 0, then there exist Gn

::::J

B, Gnt such that (20)

Proof. The basic formula here is (16) of §2.4:

T B = lim

t (s + DB

0

8 s ),

(21)

sllO

applied to a sequence Sk H O. Let P.k = p.PSk ' Then for each k?: 1, we have by part (a): there exist K kn C B, K kn j as nj such that

which means (22) Let K n = u~~ 1 K kn : then K n c B, K nj, and it is clear from (22) that

Therefore, we have by (an analogue of) Lemma 1 in §3.2: ~=~t~+~o~=~t~t~+~o~ k

k

n

= lim t lim t (Sk + D Kn esJ = lim t T Kn , n k 0

This proves (19) and it is important to see that the additional condition p.(B) = 0 is needed for (20). Observe that T B = DB unless X 0 E B; hence under the said condition we have PI'{TB = DB} = 1. It follows from this remark that T G = D G for an open G, because if X a E G we have T G = 0 (which is not the case for an arbitrary B). Thus we have by part (a):

o

95

3.3. Hitting Times

Remark. The most trivial counterexample to (20) when J.l(B) i= 0 is the case of uniform motion with B = {O}, J.l = 8 0 , Under pI', it is obvious that T B = 00 but T G = 0 for each open G :J B. Another example is the case of Brownian motion in the plane, if B = {x} and J.l = 8 x ; see the Example in §3.6.

As a first application of Theorem 8, we consider the "left entrance time" and "left hitting time" of a Borel set B, defined as follows:

Dii(w) Tii(w)

= inf{t ~ 0IXt-(w) E B}, inf{t > 0IXt-(w)

=

E

B}.

(23)

We must make the convention X o - = X o to give a meaning to DB' Theorem 9. We have almost surely:

Dc = DG ; Dii ~ DB;

Tc = TG ;

Tii ~ T B·

Proof. Since the relation (21) holds for Tii and Dii as well, it is sufficient to consider the D's. If X t E G and G is open, then there is an open set G1 such that 0 1 c G and X t E G 1. The right continuity of paths then implies that for each W there exists bo(w) > 0 such that Xt+o E 0 1 for 0 < 15 ::;; 15 0 and so Xt+o- E 0 1 , This observation shows Dc ::;; DG • Conversely if t > 0 and X t - E G, then there exist t n it t such that X tn E G; whereas if X o - E G then X o E G. This shows DG ::;; Dc and consequently DG = Dc. For a general B, we apply the second part of (15) to obtain Gn :J B such that DGn i DB' PI'-a.s. Since Dii ~ DCn = DGn , it follows that Dii ~ DB' PI'-a.s.

o

We can now settle a point remarked in §3.1, by applying Theorem 9 with B

= {o}.

Corollary. For a Hunt process, we have almost surely:

1ft < (: X t -

E

E.

(24)

n,

This implies: on the set {t < the closure of the set of values USE [O,t) X(s, w) is a compact subset of E. We close this section by making a "facile generalization" which turns out to be useful, in view oflater developments in §3.4. Definition. A set A c Eo is called nearly Borel iff for each finite measure J.l on there exist two Borel sets B 1 and B 2 , depending on J.l, such that B 1 c A e B 2 and (25)

{fa,

96

3. Hunt Process

This is equivalent to: PIl{:Jt

E

[0, oo):Xt

E

B2

-

Bd = 0.

(26)

Of course, this definition depends on the process {X t }. The class of nearly Borel sets will be denoted by Iff '. It is easy to verify that this is a O"-field. It is included in Iff-, the universally measurable O"-field, because (26) implies

One can describe Iff' in a folksy way by saying that the "poor" Hunt process {Xt} cannot distinguish a set in Iff' from a set in Iffo. It should be obvious that the hitting time of a nearly Borel set is optional, and the approximation theorems above hold for it as well. [We do not need "nearly compact" or "nearly open" sets!] Indeed, if A E Iff' then for each j1 there exists B E Iffo such that DB = DA' PIl-a.s. Needless to say, this B depends on j1. A function 1 on Eo to [ - 00, + 00] is nearly Borel when 1 E Iff'. This is the case if and only if for each j1, there exist two Borel functions 11 and 12, depending on j1, such that 11 sis 12 and (27) It follows that we may replace 11 or 12 in the relation above by 1. We shall

see in the next section that all excessive functions are nearly Borel.

3.4. Balayage and Fundamental Structure Let {Xt,~} be a Hunt process, where {~} is as specified in §3.1. For 1 E Iff+, rx 2 0, and optional T, we define the operator P~ as follows: (1)

°

We write P T for P~. If we use the convention that X = i} and 1(8) = for each 1, then we may omit "T < 00" in the expression above. We shall frequently do so without repeating this remark. The following composition property is fundamental. Recall that if Sand T are optional relative to {~}, so is S + To Os: see Exercise 7 of §2.3. CD

Proposition 1. We have

(2)

97

3.4. Balayage and Fundamental Structure

Proof. Since X T(W)

= X(T(w),w),

It follows that for f

E

P'SP;'f

we have

Mf + :

=E·{e-~SP;'f(Xs)}

= =

=

E·{e-~S[e-~Tf(XT)J

E·{e-~SEXs[e-~Tf(XT)]}

o8 s }

E·{e-~(s+ToesY"(Xs+Toes)}'

D

Here and henceforth we will adopt the notation E·(· .. ) to indicate the function x ~ E X (, • '). Two important instances of (2) are: when S = t, a constant; and when S and T are hitting times of a nearly Borel set A. In the latter case we write P A for P;'A • We shall reserve the letter A for a nearly Borel set below. Definition. A point x is said to be regular for the set A iff

r{TA = O} = 1.

(4)

The set of all points in Eo which are regular for A will be denoted by A r ; and the union A u A r is called the fine closure of A and denoted by A *. The nomenclature will be justified in §3.5. According to the zero-or-one law (Theorem 6 of §2.3), x is not regular for A iff PX{TA = O} = 0 or r{TA > O} = 1. In this case we say also that the set A is thin at the point x. Let A denote the topological closure of A, then if

x ¢ A, a path starting at x must remain for some time in an open neighborhood of x which is disjoint from A, hence x cannot be regular for A. Thus A * c A. Since in general the set {T A = O} belongs to :?o = :?o and not to :?g, the function x ~ r {T A = O} belongs to Iff- rather than Iff. This is a nuisance which will be ameliorated.* Observe that (4) is equivalent to (5)

for each ()( > 0; indeed r{TA = O} may be regarded as Finally if x EAr, then for any f we have

lim~ioo

P Al(x).

Theorem 2. For each x and A, the measure Pix,') is concentrated on A*. In

other words,

*See Exercise 3 after §3.5.

98

3. Hunt Process

Proof. We have by definition (A*Y = A C n (Art Ify = 1. It follows that

E

(Ary, then PY{TA > O}

PX{TA < 00; X(T A) E (A*Y) ::;; EX{TA < 00; X(T A) E A C ; pX(TA)[TA > OJ}.

(6)

Applying the strong Markov property at T A, we see that the right member of (6) is equal to the r-probability of the set

where T A eTA means the "time lapse between the first hitting time of A and the first hitting time of A thereafter." If XT)w) fj; A and T A eT)w) > 0 then the sample function X(',w) is not in A for TE [TA(w), TA(w) + TAo eT)w)), a nonempty interval. This is impossible by the definition of T A' Hence the right member of (6) must be equal to zero, proving the assertion of the theorem. D 0

0

The following corollary for a = 0 is just (21) of §2.4, which is true for T B as well as DB' Corollary. If A is a closed set, then P~(x, .) is concentrated in A for each a and each x.

~

0

The operator P A(X, .) corresponds to what is known as "balayage" or "sweeping out" in potential theory. A unit charge placed at the point x is supposed to be swept onto the set A. The general notion is due to Poincare, for a modern analytical definition see e.g. Helms [1]. Hunt was able to identify it with the definition given above, and that apparently convinced the potential theorists that "he's got something there"! We interrupt the thrust of our preceding discussion by an obligatory extension of measurability. According to Theorem 5 of §2.3, for Y E bg;hence also for Y E .'1'-:;:', the function x -4 EX(y) is universally measurable, namely in g- defined in (18) of §2.3. In particular P~1 is so measurable for A E g. We have therefore no choice but to enlarge the class of functions considered from g to g-. This could of course have been done from the outset but it is more convincing to introduce the extension when the need has arisen. From here on, the transition probability measure Pt(x, .) is extended to rff-. It is easy to see that for each A E rff-,

belongs to fjJ x rff-; alternately, (t,x) -4 Ptf(x) is in [lJ x g- for each fE brffor g-:;:.. It follows that Vaf is in g- for such an f and a ~ O. Finally, an aexcessive function is defined as before except that f E g-, rather than f E g

99

3.4. Balayage and Fundamental Structure

as previously supposed. A universally measurable function is sandwiched between two Borel measurable functions (for each finite measure It on 0"0) and this furnishes the key to its handling. See the proof of Theorem 5 of §2.3 for a typical example of this remark. In what follows we fix the notation as follows: f and g are functions in 0":; x E Eo; IX ~ 0; T is optional relative to {~n; A is a nearly Borel set; K is a compact set; G is an open set. These symbols may appear with subscripts. We begin with the formula: (7)

which is derived as follows. Making the substitution t = T + u, we transform the right member of (7) by the strong Markov property into OO

E·{e-aT[fo e-aUf(Xu)dU] 0 8T} = E·{e-aTEXr[fooo e-aUf(Xu)dU]}

= E·{e- aT V'1(X T)}, which is the left member of (7). It follows at once that (8)

On {T = oo} we have P'yVaf = 0 by our convention. Note also that when T is a constant (8) has already been given in §2.1. We use the notation l.£ for the support of f, namely the set

{xEEolf(x»O}. Theorem 3. (a) (b)

(c)

P~V'1 ~ p~vaf =

vaf; vaf, if If we have

11 c

A; vaf

on

~

vag

(9)

l.£, then (9) is true everywhere.

Proof. Assertion (a) is obvious from (8) with T = T A; so is (b) if we observe thatf(X t ) = 0 for t < T A so that the first term on the right side of(8) vanishes. To prove (c) we need the approximation in Theorem 8(b) of§3.3. Let A = given x, let K n c A such that T K n t T A , r-a.s. We have then by (7), as n ....... 00:

11;

P'tvaf(x) = EX {fT: e-a1(Xt)dt}

t W{fT: e- a1(Xt)dt}

n

= p~vaf(x).

(10)

100

3. Hunt Process

This is a fundamental limit relation which makes potentials easy to handle. By the Corollary to Theorem 2, P KjX, .) is concentrated on K m hence Uaf:s uag on K n by the hypothesis of (c), yielding

Letting n ~

00,

using (a) and (b), and (10) for both f and g, we obtain

D Assertion (c) is known as the "domination principle" in potential theory, which will be amplified after Theorem 4. Note however it deals with the potentials of functions rather than potentials of measures, which are much more difficult to deal with. We state the next few results forS, but they are true for sa, mutatis mutandis. Theorem 4. Let f

E

S. Then P Af E Sand P Af :S f. If Al cAl, then (11)

For each x, there exist compacts K n c A such that

(12) Proof. Let us begin by recording an essential property of a hitting time which is not shared by a general optional time: 'It;::: 0: T A :S t

+ TAo

(Jt;

and

T A = lim

1 (t + TAO

(Jt).

(13)

tLO

This is verified exactly like (16) of §2.4, and the intuitive meaning is equally obvious. We have from (6) and (7):

It follows from this and (13) that p~ uag E S. For f E S, we have by Theorem 9 of §3.2, f = lim k l' Uag k where !Y. > O. Hence P~f = lim k l' P~ Vag k and so P~f E sa by Proposition 2 of §3.2. Finally P Af = limaLO l' P~f; hence P Af E S by Proposition 4 of §3.2. Next if Al cAl, then it follows at once from (7) that P~, vag :S P~2 vag. Now the same approximations just shown establish (11), which includes P Af :S f as a particular case. Next, given x and A, let {K n } be as in the preceding proof. As in (10), we have

101

3.4. Balayage and Fundamental Structure

Hence by Lemma 1 of §3.2,

= lim i

P~f

k

= lim i

P~ Uag k =

lim

n

Letting IY.

!

i

lim

i

lim

i

Pit uag k

n

k

= lim i

P'tuagk(X)

p~J(x).

n

k

°

and using Lemma 1 once again we obtain (12).

Corollary. Part (c) of Theorem 3 is true

o

if vag is replaced by any excessive

function in (9). The next result is crucial, though it will soon be absorbed into the bigger Theorem 6 below. Theorem 5. Let f

E

S, and A

E

Iff·. If x EAr, then

inf f :::;; f(x) :::;; sup f· A

A

Proof. Since x EAr, f(x) = PAf(x) as noted above. It follows from Theorem 8(b) of §3.3 that for each e > 0, there exists K c A such that (14) since PX{TA = O} = 1. At the same time, we may choose K so that

by (12). Hence by the Corollary to Theorem 2, we have

PK(x,K) inf f(y):::;; PKf(x):::;; supf(y)· YEK

YEK

Since PK(X, K) ::2: p x {TK :::;; e} ::2: 1 - e by (14), it follows from the inequalities above that (1 - e) inf f(y) :::;; f(x) :::;; sup f(y) yEA

+ e.

YEA

Since e is arbitrary this is the assertion of the theorem.

o

We have arrived at the following key result. In its proof we shall be more circumspect than elsewhere about pertinent measurability questions. E S. Then almost surely t ~ f(X t ) is right continuous on [0,00) and has left limits (possibly + 00) in (0, 00]. Moreover, f E Iff·.

Theorem 6. Let f

102

3. Hunt Process

Proof. We begin by assuming that f E Sniff·. Introduce the metric p on [0,00] which is compatible with the extended Euclidean topology: p(a, b) = I(all + a) - (b/l + b)l, where 00/(1 + 00) = 1. For each E > define

°

S,(w) = inf{t > Olp(f(Xo(w)),f(Xt(w))) > E}.

An argument similar to that for the Tin (2) of §3.1 shows that S, is optional (relative to (Ji';)). It is clear that we have

Dl

{SIlk> O}

C

{wh~ f(Xt(w)) = f(Xo(w))} = A,

(15)

say. For each x put A = {y E EiJlp(f(x),f(y)) > E}. Then A E Iff·; and S, reduces to the hitting time T A under PX. It is a consequence of Theorem 5 that x ¢ A'; for obviously the value f(x) does not lie between the bounds of f on A. Hence for each x, PX{S, > O} = 1 by Blumenthal's zero-one law. Now {Sa > O} E ff, hence P'{S, > O} E Iff- by Theorem 5 of §2.3, and this implies pll{S, > O} = 1 for each probability measure /1 on lffiJ. It now follows from (15) that A E ffll since (ffll,pll) is complete. This being true for each /1, we have proved that A E ff and A is an almost sure set. For later need let us remark that for each optional S, A 8s E ff by Theorem 5 of §2.3. Our next step is to define inductively a family of optional times Til. such that the sample function t ---+ f(X t ) oscillates no more than 2E in each interval [Til., Ta+ 1)' This is analogous to our procedure in the proof of Theorem 1 of §3.1, but it will be necessary to use transfinite induction here. The reader will find it instructive to compare the two proofs and scrutinize the difference. Let us denote by the well-ordered set of ordinal numbers before the first uncountable ordinal (see e.g., Kelley [1; p. 29J). Fix E, put To == 0, T 1 == S" and for each IX EO, define Til. as follows: 0

°

(i)

if IX has the immediate predecessor

(ii)

if IX is a limit ordinal without an immediate predecessor, put Til.

IX -

1, put

= sup T p where fJ ranges over all ordinals <

IX.

p<1l.

Then each Til. is optional by induction. [Note that the sup in case (ii) may be replaced by the sup of a countable sequence.J Now for each IX EO, we have for each /1: pll {Til. < 00; lim f(X T"+t) = f(X TJ} tt 0

= Ell {Til. < 00; pX(T") [~~~ f(X t) = f(X o)]} = PIl{TIl. < (f)} (16)

103

3.4. Balayage and Fundamental Structure

by the strong Markov property and the fact that PX(A) = 1 for each x, proved above. It follows that pl'-a.e. on the set {Ta < oo}, the limit relation in the first member of (16) holds; hence T 1 ()T, > 0 by the definition of T 1 , hence T a + 1 > T a · Consider now the numbers 0

a EO. We have just shown that if ba > 0, then

Now the set 0 has uncountable cardinality. If ba > 0 for all a E 0, then the uncountably many intervals (C a + 1, Ca ), a EO, would be all nonempty and disjoint, which is impossible because each must contain a rational number. Therefore there exists a* E 0 for which ba* = 0; namely that

For each a < a*, if sand t both belong to [T a , T a +1), then p(f(Xs),f(X t )) S 28. This means: there exists Q~ with pl'(Q,) = 1 such that if W E Q~, then the sample function t ~ f(X(t,w)) does not oscillate more than 28 to the right of any t E [0, 00). It follows that the set Q; of w satisfying the latter require1 Qt/k is contained ment belongs to ff and is an almost sure set. Since in the set of w for which t ~ f(Xt(w)) is right continuous in [0,00), we have proved that the last-mentioned set belongs to ff and is an almost sure set. This is the main assertion of the theorem. In order to fully appreciate the preceding proof, it is necessary to reflect that no conclusion can be drawn as to the existence ofleft limits of the sample function in (0, 00). The latter result will now follows from an earlier theorem on supermartingales. In fact, if we write fn = f!\ n then {fn(Xt)'~} is a bounded supermartingale under each PX, by Proposition 1 of §2.1, since fn is superaveraging. Almost every sample functions t ~ fn(X t) is right continuous by what has just been proved, hence it has left limits in (0, 00 ] by Corollaries 1 and 2 to Theorem 1 of §1.4. Since n is arbitrary, it follows (why?) that t ~ f(X t ) has left limits (possibly + 00) as asserted. It remains to prove that f E S implies f E tff·. In view of Theorem 9 of §3.2, it is sufficient to prove this for Uag where a> 0 and g E btff:. For each jl, there exist g1 and gz in tff a such that g1 S g S gz and jlUa(gz - g1) = O. For each t 2: 0, we have

nk"=

It follows that

104

Thus we have for each t

3. Hunt Process ~

0: PI'-a.s.

(17)

Therefore (17) is also true simultaneously for all rational t > 0. But U~(g2 - gl) E S~ n tff, hence the first part of our proof extended to s~ shows that the function of t appearing in (17) is right continuous, hence must vanish identically in t, PI'-a.s. This means U~g E tff° as asserted. Theorem 6 is completely proved. D In the proof of Theorem 6, we have spelled out the 57-measurability of "the set of all ill such that f(X(·, ill)) is right continuous". Such questions can be handled by a general theory based on T x Q-analytic sets and projection, as described in §1.5; see Dellacherie and Meyer [IJ, p. 165. If f E S, when can f(X t) = oo? The answer is given below. We introduce the notation (18) 1ft E (0, 00]: f(X t)- = lim f(X s )' siit

This limit exists by Theorem 6, and is to be carefully distinguished from f(X t -). Note also that in a similar notation f(X t)+ = f(X t) = f(X t +). Theorem 70 Let F = {f < oo}. Then we have almost surely: for all t > DF,

(19)

where DF is the first entrance time into F. Proof. If x E F, then under r, {j(Xt),~} is a right continuous positive supermartingale by Theorem 6. Hence by Corollary 1 to Theorem 1 of §1.4, f(X t ) is bounded in each finite t-interval, and by Corollary 2 to the same theorem, converges to a finite limit as t -'> 00. Therefore (19) is true with D F = 0. In general, let F n = {j :::; n}, then F n E tff° by Theorem 6 and so D Fn is optional; also f(X(DFJ) :::; n by right continuity. It follows by the strong Markov property that (19) is true when the D F there is replaced by D Fn . Now an easy inspection shows that limn! D Fn = D F, hence (19) holds as written. D

00

It is obvious that on the set {D F > O},f(X t ) = 00 fort < DF,andf(X t )for t :::; DF; but f(X(D F)) may be finite or infinite.

=

There are numerous consequences ofTheorem 6. One of the most important is the following basic property of excessive functions which can now be easily established. Theorem 80 The class of excessive functions is closed under the minimum operation.

105

3.4. Balayage and Fundamental Structure

Proof. Let f1 other hand,

E

S, f2

E

S, then we know f1 /\f2 is superaveraging. On the

2: EX

{lim [f1(X )/\f2(X qo t

t)]}

= g{f1(X O) /\f2(X O)} = (f1/\f2)(X); by Fatou's lemma followed by the right continuity of /;(X t ) at t = 0, i = 1,2. 0 Hence there is equality above and the theorem is proved. In particular, if f E S then f /\ n E S for any n. This truncation allows us to treat bounded excessive functions before a passage to the limit for the general case, which often simplifies the argument. As another application of Theorem 6, let us note that P A1 is a-excessive for each (X 2: and each A E ,g' , hence P A1 E ,g' . Since AT = {x IP A1 = I} as remarked in (5), we have AT E ,g', A * E ,g'. The significance of these sets will be studied in the next section.

°

Exercises

1. Let f

E

b,g +, Tn

i

T. Prove that for a Hunt process

!~~ Uj'(X(Tn)) = E {uaf(X(T))!V 9'Tn}' Hence if we assume that (20) then we have lim Uaf(X(Tn)) = Uj'(X(T)). n

Remark. Under the hypothesis that (20) holds for any optional Tn i T, the function t ~ Uaf(X(t-)) is left continuous. This is a complement to Theorem 6 due to P. A. Meyer: the proof requires projection onto the predictable field, see Chung [5].

2. Let A E,g° and suppose Tn the set nn{Tn < T A < oo}:

i

TAo Then we have for each a> 0, a.s. on

lim EX(Tn){e-aTA}

=

1.

[Hint: consider g{e- aTA ; Tn < T A I9'TJ and observe that T A

E

Vn 9'Tn']

3. Let Xo be a holding but not absorbing point (see Problem 3 of§3.1). Define T

=

inf{t > O:X(t) =I- x o}.

106

3. Hunt Process

For an increasing sequence of optional Tn such that Tn ::s:: T and Tn we have

i

T

PXOLOl [Tn < T]} = O. Indeed, it follows that for any optional S, pxo{O < S < T} = O. [Hint: the second assertion follows from the first, by contraposition and transfinite induction, as suggested by J. B. Walsh.] 4. Let I U(g+)

E J\

Iff + and I 2 P KI for every compact K. Let 9 U(g-) < 00 (namely V(g) is defined). Then

12 Vg

E

Iff such that

on ~

implies I 2 Ug (everywhere). [Hint: 1+ U(g-) 2 V(g+); for each K c ~ we have I + V(g-) 2 PK(f + U(g-)) 2 PKV(g+), now approximate by compacts.]

UL.

5. Suppose V is transient (see (17) of §3.2). For the I in Problem 4 we have 12 a V"I for every a > O. [Hint: we may suppose I bounded with compact support; put 9 = 1- aU"I, then I 2 V(ag) = V"(af) on {g 2 O}. This is a true leger-demain due to E. B. Dynkin [l].J 6. Prove Theorem 6 for a Feller process by using Theorem 5 of §1.4. [Hint: show first U"I(X t ) is right continuous for I E blff +, which is trivially true for I E Co·J 7. Let I be excessive and A = {J = oo}. If I(xo) <

00,

then PA1(xo) = O.

8. The optional random variable T is called terminal iff for every t 2 0 we have T=t+Toe t

on{T>t}

and T = lim (t ItO

+ To

e t ).

Thus the notion generalizes that of a hitting time. Prove that if T is terminal, then x ~ p x{T < oo} is excessive; and x ~ EX {e - aT} is aexcessive for each a > O.

3.5. Fine Properties Theorem 5 of §3.4 may be stated as follows: if IE S, then the bounds of I on the fine closure of A are the same as those on A itself. Now if I is continuous in any topology, then this is the case when the said closure is taken in that topology. We proceed to define such a topology.

107

3.5. Fine Properties

Definition" An arbitrary set A is called finely open iff for each x E A, there exists BEg" such that x E Be A, and

PX{TBc> O}

=

1.

(1)

Namely, almost every path starting at a point in A will remain in a nearly Borel subset of A for a nonempty initial interval of time. Clearly an open set is finely open. It is immediately verified that the class of finely open sets meets the two requirements below to generate a topology: (a) (b)

the union of any collection of finely open sets is finely open; the intersection of any finite collection of finely open sets is finely open.

We shall denote this class by O. Observe that if we had required a finely open set to belong to g", condition (a) would fail. On the other hand, there is a fine base consisting of sets in g"; see Exercise 5 below. Since 0 is absorbing, the singleton {o} forms a finely open set so that 0 is an isolated point in the fine topology. If
E g". Then f is finely continuous is right continuous on [0,(0), almost surely.

if and

only

if

t

--+

f(X t )

Proof. Assume the right continuity. For any real constant c, put A

= {f < c},

A'

= {f> c}.

For each x E A, under p x we have f(X o) = f(x) < c, hence PX{limtto f(X t ) < c} = 1 by right continuity at t = O. This implies (1) with B = A. Hence A is finely open; similarly A' is finely open. Consequently f-1(U) is finely open, first for U = (c 1 , c z) where C1 < C z, and then for each open set U of (- 00, (0). Therefore f is finely continuous, namely continuous in the fine topology, by a general characterization of continuity.

108

3. Hunt Process

Conversely, let f

E

tff· and f be finely continuous. For each q E Q, put

A

=

U > q},

Then A is finely open; also A E tff· implies that T A is optional, and ({Jq is I-excessive. It follows from Theorem 6 of §3.4 that there exists (2. with P X ((2.) = I such that if WE (2. then 't:/q

E

Q: t ---+ ((Jq(X t) is right continuous.

(2)

We claim that for such an w, limswf(X(s,w)) ~f(X(t,w)) for all t 2 0. Otherwise there exist t 2 0, t n 11 t, and a rational q such that

f(X(t,w)) < q,

f(X(tn,W)) > q.

Since X(t n, w) E A and A is finely open, the point X(tn, w) is certainly regular for A, hence ({JiX(tn'w)) = I for all n. Therefore we have ({JiX(t,w)) = I by (2). But B = U < q} is also finely open and X(t,w) E B, hence by definition the point X(t, w) is not regular for Be, a fortiori not regular for A since Be::::J A. Thus ((Jq(X(t,w)) < 1. This contradiction proves the claim. A similar argument shows that lim sHt f(X(s,W))2f(X(t,w)) for all t20, PX-a.s. Hence t ---+ f(X(t, w)) is right continuous in [0,00), a.s. 0 Corollary. If f

E

sa, then f

is finely continuous.

Theorem 2. If A l and A z are in tff·, then (A l

For each A

E

U

AzY

= A'i

u A~.

(3)

tff·, we have (AT cAr;

T A' 2 T A

(4)

almost surely.

(5)

Proof. (3) is easy. To prove (4) let f = tff· {e- TA }. Then f(x) = I if and only if x EAr. If x E (AT then f(x) 2 inf f by Theorem 5 of §3.4. Hence f(x) = 1. A'

To prove (5), note first that A r E tff· so that T A , is optional. We have by Theorem 2 of §3.4, almost surely: X(T A') E (A r)* = A r u (AT = Ar

°

where the last equation is by (4). Hence T A f)TA' = by the strong Markov property applied at T A" and this says that T A cannot be strictly greater than T A ,· 0 0

109

3.5. Fine Properties

We shall consider the amount of time that the sample paths of a Hunt process spend in a set. Define for A c Ea : J A(W)

= {t

~

0\ X(t, w) E A}.

(6)

Thus for each t ~ 0, "t E J A(W)" is equivalent to "X(t, w) E A". We leave it as an exercise to show that if A E Iff·, then J iw) E flJ, a.s. Thus if m denotes the Borel-Lebesgue measure on T, then m(J A(W)) is the total amount of time that the sample function X(·, w) spends in A, called sometimes the "occupation time" of A. It follows that (7)

and consequently by Fubini's theorem for each x: (8)

Thus the potential of A is the expected occupation time. Definition. A set A in Iff· is said to be of zero potential iff U( ., A) Proposition 3. If U"'(', A)

== O.

== 0 for some a ~ 0, then U"'(', A) == 0 for all a ~ 0.

Proof. We have by the resolvent equation, for any f3:

If U"'(-,A) == U"'lI) == 0, then UIJU"'lA(") == 0, hence UIJ(·,A) == 0.

0

It follows from (8) that if A is of zero potential, then m(J A) = 0 almost surely; namely almost every path spends "almost no time" (in the sense of Borel-Lebesgue measure) in A. Proposition 4. If A is of zero potential, then A C

= Ea -

A is finely dense.

Proof. A set is dense in any topology iff its closure in that topology is the

whole space; equivalently iff each nonempty open set in that topology contains at least one point of the set. Let 0 be a finely open set and x EO. Then almost every path starting at x spends a nonempty initial interval of time in a nearly Borel subset B of 0, hence we have W{ m(J B)} > O. Since EX{m(J A)} = 0 we have EX{m(J BnAc)} > O. Thus 0 n A C is not empty for each finely open 0, and so A C is finely dense. 0 The following corollary is an illustration of the utility of fine concepts. It can be deduced from (11) of §3.2, but the language is more pleasant here.

110

3. Hunt Process

Corollary. If two excessive functions agree except on a set of zero potential, then they are identical.

For they are both finely continuous and agree on a finely dense set. The assertion is therefore reduced to a familiar topological one. From certain points of view a set of zero potential is not so "small" and scarcely "negligible". Of course, what is negligible depends on the context. For example a set of Lebesgue measure zero can be ignored in integration but not in differentiation or in questions of continuity. We are going to define certain small sets which play great roles in potential theory. Definition. A set A in tff· is called thin iff Ar = 0; namely iff it is thin at every point of Eo. A set is semipolar iff it is the union of countably many thin sets. A set A in tff· is called polar iff

(9)

The last condition is equivalent to: g{e- aTA } == PAI(x) == 0 for each For comparison, A is thin if and only if

IY.

> O.

(10) It is possible to extend the preceding definitions to all subsets of Eo, without requiring them to be nearly Borel. For instance, a set is polar iff it is contained in a nearly Borel set which satisfies (9). We shall not consider such an extension. The case of uniform motion on R 1 yields facile examples. Each singleton 1 {X n } {x o} is thin but not polar. Next, let x n !! xC(» - CfJ, then the set A = is semipolar but not thin. For a later discussion we note also that each compact subset of A is thin. From the point of view of the process, the smallness of a set A should be reflected in the rarity of the incidence set J A- We proceed to study the relations between the space sets and time sets. For simplicity we write qJ A for PAl below for a fixed IY. which may be taken as 1. A set A will be called very thin iff A E tff· and (11) sup qJ A(X) < 1.

U;:,=

XEA

Note that (11) does not imply that SUPXE E qJ A(X) < 1. An example is furnished by any singleton {xo} in the uniform motion, since qJ{xo}(xo) = 0 and lim xiixo qJ{xoix) = 1. Proposition 5. If A is very thin, then it is thin. Furthermore almost surely J A is a discrete set (namely, finite in each finite time interval).

111

3.5. Fine Properties

Proof. Recall that ({J A is a-excessive; hence by Theorem 5 of §3.4, the sup in (11) is the same if it is taken over A * instead of A. But if x E A', then ({J A(X) = 1. Hence under (11) A' = 0 and A is thin. Next, denote the sup in (11) by () so that () < 1. Define T 1 == T A and for n ~ 1: T n+ 1 = Tn + T A ()T n. Thus {Tn, n ~ I} are the successive hitting times of A. These are actually the successive entrance times into A, because X(T n) E A* = A here. The adjective "successive" is justified as follows. On {Tn < oo}, X(T n) is a point which is not regular for A; hence by the strong Markov property we must have T A ()T n > 0, namely T n+ 1 > Tn- Now we have, taking a = 1 in ({J A: 0

0

It follows that E{e- Tn }

::0;;

()n

°

1 as n -> 00. Let Tn i T co. Then E{e- Too } =

0;

consequently P{T co = oo} = 1. Therefore, we have almost surely co

JA

=

U {T,,}I{T

n

(12)


n=l

Since

T" i

00,

o

this is a discrete set in [0,00).

Theorem 6. Each semipolar set is the union of countably many very thin sets. For each A E Iff', A\A' is semipolar. Proof. For any A

E

Iff', we define for n

~

1: (13)

Then we have (14)

For each n, it is obvious that sup xEAnA(n)

1

({JAnA(n>(X)::O;; sup ({JA(x):s;I~-. xEA(n) n

Hence each A (\ A(n) is a very thin set. If A is thin, then (14) exhibits A as a countable union of very thin sets. Therefore each semipolar set by its definition is also such a union. Finally, (14) also shows that A - (A (\ A') is such a union, hence semipolar. 0 Corollary. If A is semipolar then there is a countable collection of optional times {Tn} such that (12) holds, where each Tn> 0, a.s.

112

3. Hunt Process

This follows at once from the first assertion of the theorem and (12). Note however that the collection {Tn} is not necessarily well-ordered (by the increasing order) as in the case of a very thin set A. We summarize the hierarchy of small sets as follows. Proposition 7. A polar set is very thin; a very thin set is thin; a thin set is semipolar; a semipolar set is of zero potential. Proof. The first three implications are immediate from the definitions. If A is semipolar, then J A is countable and so m(J A) = 0 almost surely by Theorem 6. Hence U(', A) == by (8). This proves the last implication. 0

°

For Brownian motion in R 1, each Borel set of measure zero is of zero potential, and vice versa. Each point x is regular for {x}, hence the only thin or semipolar set is the empty set. In particular, it is trivial that semipolar and polar are equivalent notions in this case. The last statement is also true for Brownian motion in any dimension, but it is a deep result tantamount to Kellogg-Evans's theorem in classical potential theory, see §4.5 and §5.2. Let us ponder a little more on a thin set. Such a set is finely separated in the sense that each of its points has a fine neighborhood containing no other point of the set. If the fine topology has a countable base (namely, satifies the second axiom of countability), such a set is necessarily countable. In general the fine topology does not even satisfy the first axiom of countability, so a finely separated set may not be so sparse. Nevertheless Theorem 6 asserts that almost every path meets the set only countably often, and so only on a countable subset (depending on the path). The question arises if the converse is also true. Namely, if A is a nearly Borel set such that almost every path meets it only countably often, or more generally only on a countable subset, must A be semipolar? Another related question is: if every compact subset of A is semipolar, must A be semipolar? [Observe that if every compact subset of A is polar, then A is polar by Theorem 8 of §3.3.] These questions are deep but have been answered in the affirmative by Dellacherie under an additional hypothesis which is widely used, as follows. Hypothesis (L). There exists a measure (0 on .g., which is the sum of countably many finite measures, such that if (o(A) = then A is of zero potential. In other words, U(x, . ) « (0 for all x.

°

It follows easily that if we put ( = (oU a for any (1, > 0, then A is of potential zero if and only if ((A) = O. Hence under Hypothesis (L) such a measure exists. It will be called a reference measure and denoted by ( below. For example, for Brownian motion in any dimension, the corresponding Borel-Lebesgue measure is a reference measure. It is trivial that there exists a probability measure which is equivalent to (. We can then use P~ with ( as the initial distribution.

113

3.5. Fine Properties

We now state without proof one of Dellacherie's results; see Dellacherie

[1]. Theorem 8. Assume Hypothesis (L). Let A E S" and suppose that almost surely the set JAin (6) is countable. Then A is semipolar.

The expediency of Hypothesis (L) will now be illustrated. Proposition 9. If f1 and f2 are two excessive functions such that f1 S f2 ~-a.e., then f1 S f2 everywhere. In particular the result is true with equalities replacing the inequalities. If f is excessive and SEo f d~ = 0, then f == O.

This is just the Corollary to Proposition 4 with a facile extension. Proposition 10. Let A E S·. Under (L) there exists a sequence of compact subsets K n of A such that K n i and for each x: (15)

Proof. This is an improvement of Theorem 8 of §3.3 in that the sequence {K n } does not depend on x. Write ~(f) for ho f d~ below; and put for a fixed a> 0:

c = sup KcA

~(P~I).

There exists K n c A, K ni such that c = lim

~(P~)).

n

Let f = limn i P~); then c = ~(f) by monotone convergence. For any compact K c A, let g = limn i P1 nuK l. Both f and g are a-excessive and g 2: f by Theorem 4 of §3.4 extended to S"'. But ~(g) S c = ~(f), hence g = f by Proposition 9 (extended to S"'). Thus P~1 sf for each compact K c A, and together with the definition of f we conclude that \Ix: f(x) = sup KcA

P~I(x)

S

P~I(x).

(16)

On the other hand, for each x it follows from Theorem 4 of §3.4 that there exists a sequence of compact subsets Ln(x) such that lim P'Ux) l(x) = n

P~I(x).

114

3. Hunt Process

Therefore (16) also holds with the inequality reversed, hence it holds with the inequality strengthened to an equality. Recalling the definition of f we have proved that limn P':.:) = P~1. Let T Kn t S:2': TAo We have EX{e- aS } = EX{e- aTA }, hence PX{S = T A } = 1 which is (15). D The next curious result is a lemma in the proof of Dellacherie's Theorem 8. Proposition 11. Let A be a semipolar set. There exists afinite measure v on g' such that if BEg' and B c A, then B is polar if and only if v(B) = 0.

Proof. Using the Corollary to Theorem 6, we define v as follows: (17) If BEg' and v(B) = 0, then

If Be A, then J B c J A = Un {T n}1{Tn
Hence B is polar by Proposition 9 applied to trivial.

f

=

P B1. The converse is D

The following analogous result, without Hypothesis (L), is of interest. Proposition 12. Let A be a semipolar set. Then for each x, there exists F =

U:= 1 K n where the Kn's are compact subsets of A, such that PX-a.s. we have TA- F

=

00.

Proof. Replace ~ by x in (17) and call the resulting measure V X. Since this is a finite measure on a locally compact separable metric space, it is regular. Hence vX(A) = sup vX(K) KcA

where K ranges over compact sets. Let K n be a sequence of compact subsets of A such that vX(K n) increases to vX(A). Then its union F has the property that vX(A - F) = 0. The argument in the proof of Proposition 11 then shows that J A-F is empty PX-almost surely. D

13

1.3. Optional Times

Hence we have for lin < t:

and consequently

{T < t} =

U{T::;' t _!}n E~.

n=l

Conversely if (1) is true, then for t

E

[0, co):

and consequently w

1}

{

w

{T::;,t}=Dl T
o

EXAMPLES.

1. The lifetime ( defined in (14) of §1.2 is optional relative to {~f+}, because ais absorbing which implies (Q is the set ofrational numbers)

g
U

{Xy=a}E~f.

YEQn[O,t)

2. Suppose ~o contains all P-null sets and Vt: P{T = t} = 0. Then we have

Vt: {T::;' t} - {T < t}

E

~

so that optionality relative to {~} is the same as relative to {~+}. 3. Consider a Poisson process with left continuous paths and T = first jump time. Then T is optional relative to {~f+} but not to {~f}, because

{T = t} ¢ ~n = ~f-]. The reader is supposed to be familiar with elementary properties of the Poisson process which are "intuitively obvious", albeit sometimes tedious to establish rigorously. From now on we fix the family {~, t E [0, co]} and call T an "optional time" or "strictly optional time" according as it is optional relative to {~+} or {~}. We shall prove most results for the former case but some of them extend to the latter. Proposition 2. If Sand T are optional, then so are

S 1\ T,

SvT,

S+ T.

115

3.5. Fine Properties

Exercises

1. Prove that

~.

is a (J-field and that

~. c ~-.

2. Prove that if A is finely dense, then A r = Eo. Furthermore for almost every w, the set JA(w) is dense in T. [Hint: consider P{e- TA }.] 3. Suppose that Hypothesis (L) holds. Then for a :2: 0, each a-excessive function is Borelian. If A E ~, then Ar E ~. For each t :2: 0, x -> px {T B :::::; t} is Borelian. [Hint: the last assertion follows from the first through the Stone-Weierstrass theorem; this is due to Getoor.] 4. Suppose that for some a > 0, all a-excessive functions are lower semicontinuous. Then Hypothesis (L) holds. [Hint: consider AU~ where A = 2 - n8xn and {x n } is a dense set in E.]

In

°

5. Let AcE and A be a fine neighborhood of x. Then for a > there exists an a-excessive function

0. If :Y is a topology on Eo which renders all a-excessive functions continuous, then :Y is a finer topology than the fine topology, namely each finely open set is open in:Y. Thus, the fine topology is the least fine topology rendering all a-excessive functions continuous. [Hint: consider the

X(t, w)). [Hint: take x = 0, A = {O} in the Brownian motion in R i . See §4.2 for a proof that E {Or]

°

9. Let A be a finely open set. Then for almost every w, X(t, w) E A implies that there exists 6(w) > such that X(u, w) E A for all U E [t, t + 6(w)). Here t is generic. The contrast with Problem 8 is interesting. [Hint: suppose A E~' and let
°

10. If A is finely closed, then almost surely J A(W) is closed from the right, namely, t n H t and t n E J A(W) implies t E J A(W), for a generic t.

116

3. Hunt Process

3.6. Decreasing Limits The limit of an increasing sequence of excessive functions is excessive, as proved in Proposition 2 of §3.2. What can one say about a decreasing sequence? The following theorem is due to H. Cartan in the Newtonian case and is actually valid for any convergent sequence of excessive functions. It was proved by a martingale method by Doob, see Meyer [2]. The proof given below is simpler, see Chung [4]. Theorem 1. Let fn

E

S and lim fn = f. Then f is superaveraging and the set 00 everywhere, then

{f > ]} is semipolar. In case J <

{f > J + G}

is thin

(1)

for every G > O. Proof. For each t ~ O,fn ~ Ptfn; letting n --+ 00 we obtainf ~ Ptf by Fatou's lemma. Hence f is superaveraging. For each compact K, fn ~ PKfn by Theorem 4 of §3.4; letting n --+ 00 we obtain f ~ PKf as before. Now let A denote the set in (1) and let K be a compact subset of A; then by the Corollary to Theorem 2 of §3.4 we have f > J + G on the support of PK(X, .) for every x. Therefore, since J < 00 on A, f ~ PKf ~ PdJ + G) = PKJ + GP KI

and consequently (2)

Both PKJ and P KI are excessive by Theorem 4 of §3.4. Letting t we obtain

t 0 in (2), (3)

For a fixed x there exists a sequence of compact subsets K~ of A such that PK](x) i PAJ and another sequence K~ such that P K;;I(x) i PAl(x); by (12) of §3.4. Taking K n = K~ u K~ we see that

Using this sequence of K n in (3) we obtain (4) If J < 00 everywhere this relation implies that A' = 0; for otherwise if x E A' it would read J(x) ~ J(x) + G which is impossible. Hence in this case

117

3.6. Decreasing Limits

°

A is thin. Letting c 1 through a sequence we conclude that {f >!} is semipolar. In the general case, let f~m) = fn /\ m, pm) = f /\ m. We have just shown that {f(m) > J(m) + c} is thin for each c > 0. Notice that pm) is superaveraging and J /\ m ~ J(m). It follows that 00

{j>J+c}=

U {j/\m > (J/\m) + c} m=l 00

c

U {jim) > j<m) + c} m=l

and therefore {f > J + c} is semipolar. Hence so is {j >!}.

o

Corollary. If the limit function f above vanishes except on a set of zero potential, then it vanishes except on a polar set.

Proof. Let c > 0, A = {j ~ c} and K be a compact subset of A. Then we have by the Corollary to Theorem 2 of §3.4:

(5)

Hence PKI vanishes except on a set of zero potential, and so it vanishes on a finely dense set by Proposition 4 of §3.5. But PKI being excessive is finely continuous, hence it vanishes identically by Corollary to Proposition 4 of §3.5. Thus K is polar. This being true for each compact K c A, A is polar by a previous remark. 0 The corollary has a facile generalization which will be stated below together with an analogue. Observe that the condition (6) below holds for an excessive f. Proposition 2. Let f

E

tff:

and suppose that we have for each compact K: (6)

If {j =I O} is of zero potential, then it is polar. If {j = CD} is of zero potential, then it is polar. Proof. The first assertion was proved above. Let K c {j = CD}, then we have as in (5):

Hence P K 1 vanishes on {j < CD}. The rest goes as before.

o

118

3. Hunt Process

An alternative proof of the second assertion in Proposition 2, for an excessive function, may be gleaned from Theorem 7 of §3.4. For according to that theorem, almost surely the set {t:f(X(t» = oo} is either empty or of the form [0, D F ) or [0, D F ], where D F is the first entrance time of U < oo}. If U = oo} is of zero potential then U < oo} is finely dense by Proposition 4 of §3.5, which implies that D F = because {t If(X(t)) = oo} cannot contain any nonempty interval (see Exercise 2 of §3.5). A cute illustration of the power of these general theorems is given below.

°

EXAMPLE. Consider Brownian motion in R 2 . We have Pt(x, dy) = Pt(x - y) dy where p(x) =_1 exp(_llxI12) t 2m 2t'

Well known convolution property yields Pspb)

=

fR2 Ps(x -

y)pt(y)dy

=

Ps+t(x).

Define for a > 0:

this is just ulX(x) in (10) of §3.7 below. For each s ?: 0, we have

from which it follows that f is a-excessive. It is obvious from the form of Pt(x) thatf(x) is finitefor x =1= 0, andf(o) = + 00. Thus U = oo} = {o}. Since Pk, {o}) = for each t, {o} is of zero potential. By the second assertion of Proposition 2 applied to the semigroup (e-lXtp t), we conclude that {a} is a polar set. Therefore every singleton is polar for Brownian motion in R 2 , and consequently also in R d , d ?: 2, by considering the projection on R 2 . We shall mention this striking fact later on more than one occasion. For a = we have f == 00 in the above. It may be asked whether we can find an excessive function to serve in the preceding example. The answer will be given in Theorem 1 and Example 2 of the next section.

°

°

The next theorem deals with a more special situation than Theorem 1. It is an important part of the Riesz representation theorem for an excessive function (see Blumenthal and Getoor [1], p. 84).

Theorem 3.Let B n E If' and B n 1; let Tn = T Bn • Let f E Sand 9 = limn PTJ· Then 9 is superaveraging and 9 = {j on {g < oo} (\ (nn B~Y.

119

3.6. Decreasing Limits

Proof. We have by Fatou's lemma,

Ptg

= pt(lim PTJ) slim ptPTJ slim PTJ = g n

n

n

since PTJ is excessive by Theorem 4 of §3.4. Thus g is superaveraging. If g(x) < 00, then there exists no such that PTJ(x) < 00 for n 2 no. We have by Proposition 1 of §3.4, ptPTJ(x)

= 2

But for any A

E

+ Tn W{f(X(t + Tn EX{f(X(t

0

0

et); Tn < oo} e t < Tn < t );

oo}.

(7)

g'

(8)

This relation says that the first hitting time on A after t is the first hitting time of A after 0, provided that A has not yet been hit at time t. It follows from (7) that

and consequently by subtraction: (9)

Now under px, {f(X(Tn)), :Ji7Tn , n 2 l} is a supermartingale, and {Tn S t} :Ji7 Tn. Therefore, by the supermartingale inequality we have EX{f(X(Tn)); Tn S t} 2 W{f(X(Tn+ 1 )); Tn S t} 2 W{f(X(T n+ 1)); T n+ 1 S t}.

E

(10)

Thus the right member of (9) decreases as n increases. Hence for each k 2 no we have for all n 2 k: (11) If n ---+ 00, then ptPTJ(x) ---+ Ptg(x) by dominated convergence, because PTJ(x) s PTJ(x), and ptPTJ(x) s PTJ(x) < 00. It follows from (11) that (12)

Ifx ¢ B~, thenPX{Tk > O} = 1. Letting t 1Oin(12)weobtaing(x) - g(x) = O. This conclusion is therefore true for each x such that g(x) < 00 and x¢ 1 B~, as asserted. 0

nk'=

120

3. Hunt Process

Two cases of Theorem 3 are of particular interest.

Un

Case 1. Let Gn be open, Gn- 1 C Gn and Gn = E. For each x E E, there (G~Y = {8}. If we put exists n such that x E Gn -1 so that x ¢ (G~r Hence f(8) = 0 as by usual convention, then

lim P T G nJ = g =

g

on

nn

U<

oo}.

(13)

Un K n =

E. Then nn(K~Y =

n

We can also take K n compact, K n- 1 C K~ and {8} and (13) is true when Gn is replaced by Kn-

nn

Case 2. For a fixed Xo =I 8 let Gn be open, Gn :::> Gn+1 and Gn = {x o}. For each x =I xo, there is an n such that x ¢ G~. Hence for any such x for which f(x) < 00, we have

lim PTGJ(x) = g(x) = g(x). n

On closer examination the two cases are really the same! In a sense the next result is an analogue of Theorem 3 when the optional times Tn are replaced by the constant times n. Theorem 4. Let f

E

S and suppose that hex) = lim

! PJ(x) <

(14)

00

t~OCJ

for all x. Then h

=

Pth for each t

;?:

0; and

f =h

+ fo

(15)

where fo is excessive with the property that

(16)

lim PJo == 0. t~OCJ

Proof. For each x, the assumption (14) implies that there exists to = to(x) such that Ptof(x) < 00. Since PJ ::; Ptof for t ;?: to, and Ps(Ptof)(x) ::; ptJ(x) < 00 for s ;?: 0, the following limit relation holds by dominated convergence, for each s ;?: 0: Psh = ps(lim PrJ) = lim psPJ = lim Ps+J = h. t-(X)

t~oo

Put fo

=

f - h, so that fo(x) =

00

if f(x)

(17)

(-00

= 00.

Then we have for t

;?:

to:

PJo = PJ - Pth = PJ - h,

and it follows that fo is excessive. Letting t ~

00

in (18) we obtain (16).

(18)

D

121

3.6. Decreasing Limits

This theorem gives a decomposition of an excessive function into an "invariant" part h and a "purely excessive" part fo which satisfies the important condition (6) in Theorem 6 of §3.2. It is an easy and rather emasculated version ofthe Riesz decomposition in potential theory. The reader is welcome to investigate what happens when the condition (14) is dropped. A noteworthy instance of Theorem 4 is when f = PAl where A E tff'0. In this case we have h(x) = lim PtPAl(x) = lim PX{t (-+00

Since t

+ TA

0

()t

+ TA

()t < oo}.

0

(19)

t-+oo

increases with t, we may write (19) as (20)

A little reflection shows that the set within the braces in (20) is exactly the

set of W for which J A(W) is an unbounded subset of T, where J A is defined in (6) of §3.5.

Definition. The set A is recurrent iff the probability in (19) is equal to one for every x E E; and transient iff it is equal to zero for every x E E. Of course this is not a dichotomy in general, but for an important class of processes it indeed is (see Exercise 6 of §4.1). Clearly if A is recurrent then PAl == 1 (on E). The converse is true if and only if(P t ) is strictly Markovian. The phenomenon of recurrence may be described in a more vital way, as follows. For each A E tff'0 we introduce a new function: Liw)= sup{t ~ OIX(t,w) E A}

(21)

where sup 0 = 0 as by standard and meaningful convention. We call LA the "quitting time of A", or "the last exit time from A". Its contrast to the hitting time T A should be obvious. To see that LA E g;-, note that for each w E Q: {LA(w) S t} == {X(s, w) ~ A for s > t} == {t

+ TA

0

()t(w)

=

oo}.

(22)

It follows that 00

{LA < oo}=

U {TA

0

()n

= oo};

(23)

0

()n

< oo}.

(24)

n=l

n {TA 00

{LA = oo}=

n=l

Thus A is recurrent [transient] if and only if LA = 00 [ < 00] almost surely. This may be used as an alternative definition.

122

3. Hunt Process

It is clear from the first equivalence in (22) that LA is not optional; indeed {LA:s; t} E a(X., s E (t, (0)), augmented. Such a random time has special properties and is a kind of dual to a hitting time. We shall see much of it in §5.1. Exercises 1. Let f E g + and T = T A where A we have

E

g. Suppose that for each x and t > 0,

f(x) ~ EX{I(Xt)I{t
00,

then {I(X t)1{t< T)' :#'n r} is a supermartingale.

2. In the notation of Case 1 of Theorem 3, show that for each compact K c E, we have g = PKeg on {g < oo}. 3. Under Hypothesis (L), let F be a class of excessive functions. Then there exists a decreasing sequence {In} in F such that if u = limn! fn we have u ~ infjEF f ~ u. [Hint: choose fn decreasing and limn ~(f,.I(I + fn)) = infjEF ~(f/(I + f)).]

3.7. Recurrence and Transience A Hunt process will be called recurrent iff any, hence every, one of the equivalent conditions in the following theorem is satisfied. We use the convention below that an unspecified point, set, function is of, in, on E, not Eo. Thus a function is said to be a constant ifit is a constant on E; its value at a, when defined, is not at issue. Neverthless, the proof of Theorem 1 is complicated by the fact that it is not obvious that assertion (1) holds prima facie under each of the conditions (i) to (iv) there. The additional care needed to avoid being trapped by a is well worth the pain from the methodological point of view. Recall that' = T{o}' Theorem 1. The following four propositions are equivalent for a Hunt process, provided that E contains more than one point.

(i) (ii) (iii) (iv)

Each excessive function is a constant. For each f E g~, either Uf == or Uf == 00. For each B in g' which is not thin, we have PBI == 1. For each B in g', either PBI == 0 or PBI == 1. Namely, each nearly Borel set is either polar or recurrent as defined at the end of §3.6.

°

Moreover, any of these conditions implies the following: Vx:

rg =

cD} = 1.

In other words, (P t) is a strictly Markovian semigroup on E.

(1)

123

3.7. Recurrence and Transience

Proof (i) => (ii). Since Uf is excessive, it must be a constant, say c, by (i). Suppose c =1= 00. We have for t ~ 0: (2)

Thus S~ f(X s ) ds is integrable under EX, hence finite PX-almost surely. We now prove that (i) implies (1), as follows. Let Xl and Xz be two distinct points. Then there exist open sets 0 1 and Oz with disjoint closures such that Xl E 01' X z E Oz. Both Pall and P o2 1 are excessive functions which take the value one, hence both must be identically equal to 1 by (i). Put

and for n

~

1:

Since X(To.)E 0 1 c (Oz)", we have T 1 < T z on {T 1 < oo}. The same argument then shows that the sequence {Tn} is strictly increasing. Let limn Tn = T; then on {T < oo}, X(T - ) does not exist because X(T n) oscillates between 0 1 and Oz at a strictly positive distance apart. This is impossible for a Hunt process. It follows that almost surely we have T = 00, and this implies (1) because ais absorbing. Hence PtC = C for all t. We now have by (2) and dominated convergence: C

= lim PtC = lim PtUf = t~

00

o.

t-+ 00

Remark. It is instructive to see why (i) does not imply (ii) when E reduces to one point.

(ii) => (iii). Fix BE Iff" and write CfJ for PBI. By Theorem 4 of §3.6, we have CfJ = h + CfJo where h is invariant and CfJo is purely excessive. Hence by Theorem 6 of §3.2, there exists fn E Iff ~ such that CfJo = limn 'I Ufn- Since Ufn :::;; CfJo :::;; 1, condition (ii) implies that Ufn == 0 and so CfJo == O. Therefore CfJ = PtCfJ for every t ~ O. If B is not thin, let Xo E B r so that CfJ(x o) = 1. Put A = {x E EI CfJ(x) < I}. We have for each t ~ 0:

Since Pt(xo, Eo) = 1, the equation above entails that Pt(xo, A) = O. This being true for each t, we obtain UI A (xo) = O. Hence condition (ii) implies that Ul A == O. But A is finely open since CfJ is finely continuous. If x E A, then UI A (x) > 0 by fine openness. Hence A must be empty and so CfJ == 1.

124

3. Hunt Process

(iii) => (iv). Under (iii), P B 1 == 1 for any nonempty open set B. Hence (1) holds as under (i). If B is not a polar set, there exists Xo such that P B 1(xo) = 6 > 0. Put

(3) then A is finely open. Since Xo by (iii). Now we have

E

A,

Xo E

A r by fine openness. Hence PAl

== 1 (4)

where the first inequality is by Theorem 4 of §3.4, the second by Theorem 2 of §3.4, and the third by (3) and the fine continuity of PBl. Next, we have for t ;;::: 0, PX-almost surely: PX{TB <

00

I~}

= PX{TB:s:: =

1 {TB:5tj

tl~}

+ PX{t <

TB <

+ l{t
00

I~}

(5)

00].

This is because {TB:s:: t} E fFt + = fFt ; and T B = t + T B 8t on {t < T B} (see (8) of §3.6), so that we have by the Markov property: 0

PX{t < T B <

00

IfFt}

= PX{t < T B; t + T B 8t < = l{t
00

IfFt}

As t -> 00, the first member in (5) as well as the first term in the third member converges to 1{TB < <Xl}. Hence the second term in the third member must converge to zero. Now we have X(t) E E for t < 00 and therefore pX(tl[TB < ooJ;;::: 6/2 by (4). It follows that limt.... <Xl l{t (i). Let f be excessive. Iff is not a constant, then there are real numbers a and b such that a < b, and the two sets A = {f < a}, B = {f> b} are both nonempty. Let x E A, then we have by Theorems 4 and 2 of §3.4:

since f ;;::: b on B* by fine continuity. Thus P B 1(x) < 1. But B being finely open and nonempty is not polar, hence P B 1 == 1 by (iv). This contradiction proves that f is a constant. D Finally, we have proved above that (i) implies (1), hence each of the other three conditions also implies (1), which is equivalent to Pt(x, E) = 1 for every t ;;::: 0 and x E E.

125

3.7. Recurrence and Transience

Remark. For the argument using (5) above, cf. Course, p. 344. EXAMPLE 1. It is easy to construct Markov chains (Example I of §1.2) which are recurrent Hunt processes. In particular this is the case if there are only a finite number of states which communicate with each other, namely for any two states i and j we have Pii =J= 0 and Pji =J= O. Condition (ii) is reduced to the following: for some (hence every) state i we have

SOW p;;(t)dt

(7)

= 00.

Since each i is regular for {i}, the only thin or polar set is the empty set. Hence condition (iv) reduces to: p i {1(j} < oo} = 1 for every i andj. Indeed, (8)

in the notation of (21) of §3.6. For a general Markov chain (7) and (8) are equivalent provided that all states communicate, even when the process is not a Hunt process; see Chung [2].

EXAMPLE 2. For the Brownian motion in R d , the transition probability kernel Pt(x, dy) = Pt(x - y) dy where

For rx

;?:

0, define (10)

Then the resolvent kernel U~(x, dy) = u~(x - y) dy. For d = 1 or d = 2, we have u(z) = 00. Hence by the Fubini-Tonelli theorem, U(x,B)

= SB u(x - y)dy = 0 or

00

according as m(B) = 0 or m(B) > O. It follows that condition (ii) of Theorem 1 holds. Therefore, Brownian motion in R 1 or R 2 is a recurrent Hunt process. Since each nonempty ball is not thin, the sample path will almost surely enter into it and returns to it after any given time, by condition (iv). Hence it will almost surely enter into a sequence of concentric balls shrinking to a given point, and yet it will not go through the point because a singleton is a polar set. Such a phenomenon can happen only if the entrance times into the successive balls increase to infinity almost surely, for otherwise at a

126

3. Hunt Process

finite limit time the path must hit the center by continuity. These statements can be made quantitative by a more detailed study of the process. Note: Whereas under recurrence U1 B ¥= 0 implies P B 1 == 1, the converse is not true in general. For Brownian motion in R 2 , the boundary circle C of a nenempty disk is recurrent by the discussion above, but U1 c == 0 because m(C) = O.

Turning in the opposite direction, we call a Hunt process "transient" iff its semigroup is transient as defined at the end of §3.2. However, there are several variants of the condition (17) used there. It is equivalent to the condition that E is "u-transient", namely it is the union of a sequence of transient sets. Hence the condition is satisfied if each compact subset ofE is transient, because E is u-compact. However, to relate the topological notion of compactness to the stochastic notion of transience some analytic hypothesis is needed. Consider the following two conditions on the potentials (resolvents) of a Hunt process:

"If E tS'"-r: :Ie,( > 0 such that

Uf is lower semi-continuous;

"If E tS'"-r: uaf is lower semi-continuous.

(11)

(12)

Observe that there is no loss of generality if we use only bounded f, or bounded f with compact supports in (11) and (12), because lower semicontinuity is preserved by increasing limits. It is easy to see that (12) implies (11) by the resolvent equation. Recall that condition (16) of §3.2 holds for a Hunt process and will be used below tacitly. Also, the letter K will be reserved for a compact set. Theorem 2. In general, either (ii) or (iv) below implies (iii). Under the condition (11), (iii) implies (iv). Under the condition (12), (iv) implies (i).

(i) VK: U1 K is bounded. (ii) VK: U1 K is finite everywhere. (iii) :lh E tS' + such that 0 < Uh < 00 on E. (iv) VK; limt~oo P t PK l = O.

(13)

Proof. (ii) => (iii). Let K n be compact and increase to E, and put

A nk = {x E Knl U(x,K n)::; k}. The function U1 Ank is bounded above by k on the support of lAnk' hence by the domination principle (Corollary to Theorem 4 of §3.4), we have (14)

127

3.7. Recurrence and Transience

everywhere in E. Define h as follows:

f f

h=

n=lk=l

1~":'n'

(15)

k2

Clearly h ::;; 1. For each x, there exist nand k such that x E A nk by assumption (ii). Hence h > 0 everywhere and so also Uh > O. It follows from (14) that Uh ::;; 1. Thus h has all the properties (and more!) asserted in (ii). (iii)

~

(iv) under (11). Since Uh is finite, we have (16)

as in (2). Since Uh > 0 and Uh is lower semi-continuous by (11), for each K we have infxEK Uh(x) = C(K) > O. Hence by the corollary to Theorem 2 of §3.4 we have (17) The assertion (13) follows from (16) and (17). (iv) ~ (iii) and under (12) (iv) ~ (i). Let K n C K~+ 1 and Theorem 6 of §3.2 to each PK ) we have P K ) = lim

i

Un K n = E. Applying (18)

Ug nk

k

where gnk::;; k by the proof of Theorem 6 of §3.2. It follows by Lemma 1 of §3.2 that 1 = lim

i

n

P K ) = lim

i

Ugnn-

(19)

n

Define g as follows:

Then g ::;; 1 and 0 < Ug::;; 1 by (19). Next put h = Uag; then h::;; Ug::;; 1. It follows from the resolvent equation that (20)

The fact that for each x we have

128

3. Hunt Process

implies that

Hence Vh > 0 and h satisfies the conditions in (iii). Finally, under (12) h is lower semi-continuous. Since h > 0, for each K we have infxEK h(x) C(K) > 0; hence by (20):

1

Vh VI K

::;

=

C(K) ::; cxC(K)'

namely (i) is true. Theorem 2 is completely proved.

D

Let us recall that (iv) is equivalent to "every compact subset of E is transient", i.e., VK: L K <

00

(21)

almost surely.

In this form the notion of transience is probabilistically most vivid.

EXAMPLE 3. For a Markov chain as discussed in Example I above, in which all states communicate, transience is equivalent to the condition Pii(t) dt < 00 for some (hence every) state. A simple case of this is the Poisson process (Example 3 of §1.2).

Jg'

EXAMPLE 4. Brownian motion Rd, when d ~ 3, is transient. A simple computation using the gamma function yields

(22) in particular for d = 3: dy V(x,dy) = 2nx-y I I'

(23)

If B is a ball with center at the origin and radius r, we have by calculus: 2 1 r dy nr V(o, B) = 2n JB = 2'

lYf

Let

f be bounded (Lebesgue) measurable with compact support

Vf(x) = ~ r 2n JK

f(y) dy = ~ r yl 2n JKnB

Ix -

f(y) dy + ~ r yl 2n JK\B

Ix -

K, then

f(y) dy yl

Ix -

(24)

129

3.7. Recurrence and Transience

where B is the ball with center x and radius b. The first term in the last member of (24) is bounded uniformly in x by (2n)-11IfIIV(O,b) = 4- 1 1IfIW. The second term there is continuous at x by bounded convergence. It is also bounded by b-11Ifllm(K). Thus Vf is bounded continuous. In particular if f = l K , this implies condition (i) of Theorem 2. The result also implies condition (11) as remarked before. Hence all the conditions in Theorem 2 are satisfied. The case d > 3 is similar. We can also compute for d = 3 and ex ~ 0: (25)

by using the following formula, valid for any real c:

Hence for any bounded measurable f, we have V""!(x)

=

r

JR

3

1-

exp[ ~ 1 yl J2a] f(y) dy. n x - yl

(26)

For ex > 0, an argument similar to that indicated above shows that Vaf belongs to Co. This is stronger than the condition (12). For d ~ 3, ex ~ 0, ua can be expressed in terms of Bessel functions. For odd values of d, these reduce to elementary functions. Let us take this occasion to introduce a new property for the semigroup (Pt) which is often applicable. It will be said to have the "strong Feller property" ifffor each t > 0 and f E brff +, we have Pd E C, namely continuous. [The reader is warned that variants of this definition are used by other authors.] This condition is satisfied by the Brownian motion in any dimension. To see this, we write Ptf(x) =

IB Pt(x -

y)f(y) dy

+ IRd- BPt(x -

y)f(y) dy.

(27)

For any X o and a neighborhood V of x o, we can choose B to be a large ball containing V to make the second term in (27) less than e for all x in V. The first term is continuous in x by dominated convergence. Hence Ptf is continuous at X o ' We can now show that Vaf E C for ex > 0, and f E brff+ without explicit computation of the kernel va in (26). Write

130

3. Hunt Process

and observe that the integrand is dominated by Ilflle-lXt, and is continuous in x for each t > O. Hence UlXf(x n) ---+ U"f(x) as X n ---+ x by dominated convergence. Since Uf = lim o i UlXf it now follows that Uf is lower semicontinuous, without the explicit form of U. This is a good example where the semigroup (P t ) is used to advantage-one which classical potential theory lacks. Of course, a Hunt process is a lot more than its semigroup. Let us remark that for any (P t), the semi-group (P~) where P~ = e-lXtPt is clearly transient because UIX(x, E) ~ l/C(. This simple device makes available in a general context the results which are consequences of the hypothesis of transience. Theorem 9 of §3.2 is such an example by comparison with Proposition 10 there. lX

_

Exercises 1. For a Hunt process, not necessarily transient, suppose A

E Iff, B E Iff such that Be A and U(xo, A) < 00. Then if infxEB U(x, A) > 0, we have pxo {L B < oo} = 1, where L B is the last exist time from B. [Hint: define Tn = n + T B en, then U(x o, A) ?:: EXO{ U(X(Tn), A); Tn < 00 }.] 0

2. Let X be a transient Hunt process. Let Al be compact, Anl and nnAn = B. Put f = limn P A)' Prove that f is superaveraging and its excessive regularization J is equal to PBl. [Hint: f ?:: J?:: P Bl. If x rt B, then px {

by transience. Hence

0[TAn < oo]} = Px {TB < oo} J=

P B l on Be u BY which is a finely dense set.]

3. Let hE Iff + and Uh < 00. In the same setting as Problem 2, prove that if f = limn PAnUh then J = PBUh. 4. Let X be a transient Hunt process and suppose Hypothesis (L) holds with ~ a reference probability measure. Let B a compact polar set. Prove that there exists a function f E Iff+ such that ~(f) ~ 1 and f = 00 on B. [Hint: in Problem 2 take An to be open, and fn = PA)' On Be, in 1 PBl, hence ~(fn) 1 0; on B, fn = 1. Take {nk} so that ~(Lk ink) ~ 1. This is used in classical potential theory to define a polar set as the set of "poles" of a superharmonic function.] 5. Prove that (iii) in Theorem 2 is equivalent to E being O'-transient. [Hint: under (iii), {Uh > c} is transient.]

3.8. Hypothesis (B) This section is devoted to a new assumption for a Hunt process which Hunt listed as Hypothesis (B). What is Hypothesis (A)? This is more or less the set of underlying assumptions for a Hunt process given in §3.1. Unlike

131

3.8. Hypothesis (B)

some other assumptions such as the conditions of transience, or the Feller property and its generalization or strengthening mentioned in §3.7, Hypothesis (B) is not an analytic condition but restricts the discontinuity of the sample paths. It turns out to be essential for certain questions of potential theory and Hunt used it to fit his results into the classical grain. Hypothesis (B). we have

For some rx ;::: 0, any A

E

g., and open G such that A c G, (1)

This means of course that the two measures are identical. Recalling Proposition 1 of §3.4, for S = T G and T = T A, we see that (1) is true if TA= TG

+ TAo

8 TG

almost surely on {T A < oo}.

(2)

On the other hand, if rx > 0, then the following particular case of (1) (3)

already implies the truth of (2). For (3) asserts that

whereas the left member in (2) never exceeds the right member which is inf{t> T G : X(t) E A}. We will postpone a discussion of the case rx = 0 until the end of this section. Next, the two members of (2) are equal on the set {T G < T A}' This is an extension of the "terminal" property of T A expressed in (8) of §3.6. Since G ::::J A, we have always T G ~ T A. Can T G = T A? [If A c G, this requires the sample path to jump from GC into X] There are two cases to consider. Case (i). T G = T A < 00 and X(T G ) = X(T A) EAr. In this case, the strong Markov property at T G entails that TAo 8 TG = O. Hence the two members of (2) are manifestly equal. Case (ii). T G = T A < 00 and X(TG ) = X(T A) ¢ A r. Then since X(T A) E A u A r, we must have X(T G) E A\A r. Consequently TAo 8 > and the TG

left member of (2) is strictly less than the right.

°

We conclude that (2) holds if and only if case (ii) does not occur, which may be expressed as follows:

Vx: PG(x,A\A r )

=

O.

(4)

Now suppose that (2) holds and A is thin. Then neither case (i) nor case (ii) above can occur. Hence for any open Gn::::J A we have almost surely (5)

132

3. Hunt Process

Suppose first that x 13 A; then by Theorem 8(b) of §3.3 there exist open sets Gn ::::J A, Gnt such that T G n iTA' PX-a.s. Since left limits exist everywhere for a Hunt process, we have limn~oo X(TGJ = X(T A-); on the other hand, the limit is equal to X(TA) by quasi left continuity. Therefore, X(·) is continuous at T A , pX- a.s. For an arbitrary x we have PX{TA < 00; X(T A-) = X(TA )} = lim PX{t < T A < 00; X(T A-) = X(TA )} ttO

=

lim PX{t < T A ; pX(tl[TA < 00; ttO

X(T A - ) = X(T A)]} =

lim PX{t < T A ; pX(tl[TA < oo]} ttO

=

lim PX{t < T A; p A1(X t)} tto

In the above, the first equation follows from Px {T A > O} = 1; the second by Markov property; the third by what has just been proved because X(t) 13 A for t < T A; and the fifth by the right continuity of t --+ PA l(X t ). The result is as follows, almost surely: (6)

Equation (6) expresses the fact that the path is almost surely continuous at the first hitting time of A. This will now be strengthened to assert continuity whenever the path is in A. To do so suppose first that A is very thin. Then by (12) of §3.5, the incidence set J A consists of a sequence {Tn' n ?: I} of successive hitting times of A. By the strong Markov property and (6), we have for n ?: 2, almost surely on {Tn - 1 < oo}: PX{Tn < 00; X(T n-) = X(T n)} = pX(Tn-d{TA < 00; X(T A-) = X(T A)} = pX(Tn-d{TA < oo} = PX{Tn < oo}. It follows that X(T n-) = X(T n) on {Tn < oo}, and consequently

'it?: 0: X(t-) = X(t)

almost surely on {Xt E A}.

(7)

Since (7) is true for each very thin set A, it is also true for each semi-polar set by Theorem 6 of §3.5. Thus we have proved that (2) implies the truth of (7) for each thin set A c G. The converse will now be proved. First let A be thin and contained in a compact K which in turn is contained in an open set G. We will show that (4) is true. Otherwise there is an x such that PX{TG < 00; X(TG) E A} > O. Since A is at a strictly positive distance from GC, this is ruled out by (7). Hence

133

3.8. Hypothesis (B)

= O. Next, let KeG as before. Then K\K r =

Un

C n where each Cn is (very) thin by Theorem 6 of §3.5. Since Cn c K, the preceding argument yields P G(x, C n) = 0 and consequently P G(x, K\K r ) = O. By the discussion leading to (4), this is equivalent to: PG(x, A)

Finally let A E S·, A c G. By Theorem 8(b) of §3.5, for each x there exist compact sets K., K n c A, K n t such that T K n ! T A, PX-almost surely. On {T A < oo} we have T K n < 00 for all large n. Hence (8) with K = K n yields (2) as n ----t 00. We summarize the results in the theorem below. Theorem 1. For a Hunt process the following four propositions are equivalent: (i) Hypothesis (B), namely (1), is true for some ex> O. (ii) Equation (3) is true for some ex > O. (iii) For each semi-polar A and open G such that A c G, we have P G(x, A) = 0 for each x. (iv) For each semi-polar A, (7) is true. Moreover, if in (1) or (3) we restrict A to be a compact set, the resulting condition is also equivalent to the above. It should also be obvious that if Hypothesis (B) holds for some ex > 0, it holds for every ex > O. We now treat the case ex = O.

Theorem 2. If the Hunt process is transient in the sense that it satisfies condition (iii) of Theorem 2 of §3.7, then the following condition is also equivalent to the conditions in Theorem 1. (v)

Hypothesis (B) is true for

IX =

O.

Proof. Let h > 0 and Uh :::;; 1. Then applying (1) with ex = 0 to the function Uh, we obtain

On the set {TA < T G + T A eTa; T A < oo}, the integral above is strictly positive. Hence the preceding equation forces this set to be almost surely 0 empty. Thus (2) is true, which implies directly (1) for every ex ;::: O. 0

Needless to say, we did not use the full strength of (v) above, since it is applied only to a special kind of function. It is not apparent to what extent the transience assumption can be relaxed. In Theorem 9 of §3.3, we proved that for a Hunt process, T B :::;; Tii a.s. for each BE S·. It is trivial that T B = Tii if all paths are continuous. The next

134

3. Hunt Process

theorem is a deep generalization of this and will play an important role in §5.1. It was proved by Meyer [3J, p. 112, under Hypothesis (L), and by Azema [IJ without (L). The following proof due to Walsh [IJ is shorter but uses Dellacherie's theorem.

Theorem 3. Under Hypotheses (L) and (B), we have T B = Tii a.s. for each BEg'.

Proof. The key idea is the following constrained hitting time, for each BEg': SB = inf{t >

01 X(t-) = X(t) E B}.

(9)

To see that SB is optional, let {in} be the countable collection of all jump times of the process (Exercise 5 of §3.1), and A = {TB of- in}. Then A E ~(TB) by Theorem 6 of§1.3, hence SB = (TB)A is optional by Proposition 4 of§1.3. It is easy to verify that SB is a terminal time. Define cp(x) = E X { e- SH }. Then cp is I-excessive by Exercise 8 of §3.4. Now suppose first that B is compact. For < I:: < I put

U:=l

°

A

=

B

n{xlcp(x):o:; I -

I::}.

Define a sequence of optional times as follows: R 1

=

SA, and for n :2: 1:

On {R n < oo} we have X(R n) E A because A is finely closed. Hence cp(X(R n)) :0:; I - I:: and so pX(Rn){SA > O} = I by the zero-one law. It follows that R n < R n + 1 by the strong Markov property. On the other hand, we have as in the proof of Proposition 5 of §3.5, for each x and n :2: 1:

Hence R n l' 00 a.s. Put 1(w) = {t > OIX(t,w) E A}. By the definition of R m for a.e. w such that Rn(w) < 00, the set 1(w) n (Rn(w),Rn+1(w)) is contained in the set of t where X(t-,w) of- X(t,w), and is therefore a countable set. Since Rn(w) l' 00, it follows that 1(w) itself is countable. Hence by Theorem 8 of §3.5, A is a semi-polar set. This being true for each 1::, we conclude that the set B ' = B n {xlcp(x) < I} is semi-polar. Now we consider three cases on {TB < oo}, and omit the ubiquitous "a.s." (i) (ii)

X(T B-) = X(T B). Since B is compact, in this case X(T B-) E Band so Tii :0:; T B. X(T B-) of- X(T B) and X(T B) E B - B' . In this case cp(X(TB)) = I and so SB 8 TH = 0. Since Tii :0:; SB we have Tii :0:; T B. X(T B-) of- X(TB) and X(T B) E B ' . Since B ' is semi-polar this is ruled out by Theorem 1 under Hypothesis (B). 0

(iii)

135

3.8. Hypothesis (B)

We have thus proved for each compact B that Tii :s; T Bon {TB < oo}; hence Tii = T B by Theorem 9 of §3.3. For an arbitrary B E (r, we have by Theorem 8(b) of§3.3, for each x a sequence of compacts K n c B, Kni such thatr{TKn ! T B} = 1. Since T Kn = T Kn as just proved and Tii :s; T Kn for n ?: 1, it follows that r {Tii :s; T B} = 1. As before this implies the assertion ofTheorem 3. 0 Exercises

1. Give an example such that PAl = PGPAl but PA ?: PGPA is false. [Hint: starting at the center of a circle, after holding the path jumps to a fixed point on the circle and then execute uniform motion around the circle. This is due to Dellacherie. It is a recurrent Hunt process. Can you make it transient?J

2. Under the conditions of Theorem 1, show that (7) remains true if the set {Xt E A} is replaced by {X t- E A}. [Hint: use Theorem 9 of§3.3.J 3. If (7) is true for each compact semi-polar set A, then it is true for every semi-polar set. [Hint: Proposition 11 of §3.5 yields a direct proof without use of the equivalent conditions in Theorem 1.J

4. LetE= {O} u [1,00); letObeaholdingpointfromwhich the path jumps to 1, and then moves with uniform speed to the right. Show that this is a Hunt process for which Hypothesis (B) does not hold. 5. Let A be a thin compact set, Gn open, and Gn !! A. Show that under Hypothesis (B), we have under r, x ¢ A: "In: T Gn < T A,

T Gn iTA,

Thus T A is predictable under

a.s. on {T A < oo}.

px, x ¢ A.

6. Assume that the set A in Problem 5 has the stronger property that sup EX{e- TA } < 1. xEEa

Then A is polar. [Hint: use Problem 2 of §3.4.J

NOTES

ON

CHAPTER

3

§3.1. Hunt process is named after G. A. Hunt, who was the author's classmate in Princeton. The basic assumptions stated here correspond roughly with Hypothesis (A), see Hunt [2]. The Borelian character of the semigroup is not always assumed. A more general process, called "standard process", is treated in Dynkin [1] and Blumenthal and Getoor [1]. Another technical variety called "Ray process" is treated in Getoor [1]. §3.2. The definition of an excessive function and its basic properties are due to Hunt [2]. Another definition via the resolvents given in (14) is used in Blumenthal and Getoor [1]. Though the former definition is more restrictive than the latter, it yields more transparent results. On the other hand, there are results valid only with the resolvents.

136

3. Hunt Process

§3.3. We follow Hunt's exposition in Hunt [3]. Although the theory of analytical sets is essential here, we follow Hunt in citing Choquet's theorem without details. A proof of the theorem may be found in Helms [1]. But for the probabilistic applications (projection and section) we need an algebraic version of Choquet's theory without endowing the sample space Q with a topology. This is given in Meyer [1] and Dellacherie and Meyer [1]. §3.4 and 3.5. These sections form the core of Hunt's theory. Theorem 6 is due to Doob [3], who gave the prooffor a "subparabolic" function. Its extension to an excessive function is carried over by Hunt using his Theorem 5. The transfinite induction used there may be concealed by some kind of maximal argument, see Hunt [3] or Blumenthal and Getoor [1]. But the quickest proof of Theorem 6 is by means of a projection onto the optional field, see Meyer [2]. This is a good place to learn the new techniques alluded to in the Notes on §1.4. Dellacherie's deep result (Theorem 8) is one of the cornerstones of what may be called "random set theory". Its proof is based on a random version of the CantorBendixson theorem in classical set theory, see Meyer [2]. Oddly enough, the theorem is not included in Dellacherie [2], but must be read in Dellacherie [1], see also Dellacherie [3]. §3.6. Theorem 1 originated with H. Cartan for the Newtonian potential (Brownian motion in R 3 ); see e.g., Helms [1]. This and several other important results for Hunt processes have companions which are valid for left continuous moderate Markov processes, see Chung and Glover [1]. Since a general Hunt process reversed in time has these properties as mentioned in the Notes on §2.4, such developments may be interesting in the future. "Last exit time" is hereby rechristened "quitting time" to rhyme with "hitting time". Although it is the obvious concept to be used in defining recurrence and transience, it made a belated entry in the current vocabulary ofthe theory. No doubt this is partly due to the former prejudice against a random time that is not optional, despite its demonstrated power in older theories such as random walks and Markov chains. See Chung [2] where a whole section is devoted to last exits. The name has now been generalized to "co-optional" and "co-terminal"; see Meyer-Smythe-Walsh [1]. Intuitively the quitting time of a set becomes its hitting time when the sense of time is reversed ("from infinity", unfortunately). But this facile interpretation is only a heuristic guide and not easy to make rigorous. See §5.1 for a vital application. §3.7. For a somewhat more general discussion of the concepts of transience and recurrence, see Getoor [2]. The Hunt theory is mainly concerned with the transient case, having its origin in the Newtonian potential. Technically, transience can always be engineered by considering (P'!) with IX > 0, instead of (P,). This amounts to killing the original process at an exponential rate e -a" independently of its evolution. It can be shown that the resulting killed process is also a Hunt process. §3.8. Hunt introduced his Hypothesis (B) in order to characterize the balayage (hitting) operator PA in a way recognizable in modern potential theory, see Hunt [1; §3.6]' He stated that he had not found "simple and general conditions" to ensure its truth. Meyer [3] showed that it is implied by the duality assumptions and noted its importance in the dual theory. It may be regarded as a subtle generalization of the continuity of the paths. Indeed A:d:ma [1] and Smythe and Walsh [1] showed that it is equivalent to the quasi left continuity of a suitably reversed process. We need this hypothesis in §5.1 to extend a fundamental result in potential theory from the continuous case to the general case. A new condition for its truth will also be stated there.

Chapter 4

Brownian Motion

4.1. Spatial Homogeneity In this chapter the state space E is the d-dimensional Euclidean space R d, d ~ 1; iff is the classical Borel field on Rd. For A E iff, BE iff, the sets A ± B are the vectorial sum and difference of A and B, namely the set of x ± y where x E A, y E B. When B = {x} this is written as A ± x; the set 0 - A is written as - A, where 0 is the zero point (origin) of Rd. Let (P t ; t ~ 0) be a strictly stochastic transition semigroup such that for each t ~ 0, X E E and Xo E E we have (1)

Then we say (Pt) is spatially homogeneous. Temporally homogeneous Markov processes with such a semigroup constitute a large and important class. They are called "additive processes" by Paul Levy, and are now also known as "Levy processes"; see Uvy [1]. We cannot treat this class in depth here, but some of its formal structures will be presented here as a preparation for a more detailed study of Brownian motion in the following sections. We define a family of probability measures as follows. (2)

The semigroup property then becomes the convolution relation (3)

where the integral is over E and * denotes the convolution. This is valid for s ~ 0, t ~ 0, with no the unit mass at o. We add the condition: nt

~

no

vaguely as t

t 0.

(4)

138

For f

4. Brownian Motion E

tff + and t ?:: 0 we have

f f(x + y)nMy)·

PrJ(x) =

(5)

As a consequence of bounded convergence, we see that if f E be, then PrJ E bC; if f E Co, then PrJ E Co· Moreover limt!O PrJ = f for f E Co, by (4). Thus (P t ) is a Feller semigroup. Hence a spatially homogeneous process may and will be supposed to be a Hunt process. For A E tff, we define

and U"(x, A) = U"(A - x). Then U" is the o:-potential kernel and we have for f E tff+:

f

U"f(x) = f(x

+ y)U"(dy).

(6)

We shall denote the Lebesgue measure in E by m. If f E e(m), then we have by (5), Fubini's theorem and the translation-invariance of m:

f m(dx)PrJ(x) = f [f m(dx)f(x + y)]n MY ) = f [f m(dX)f(x)]n MY ) = f m(dx)f(x). For a general measure fl on tff, we define the measure flP t or flU" by (7)

The preceding result may then be recorded as Vt ?:: O.

(8)

We say m is an invariant measure for the semigroup (P t ). It follows similarly from (6) that 1

mU" = -m,

Vo: > 0;

(9)

0:

namely m is also an invariant measure for the resolvent operators (o:U"). Proposition 1. Let 0: > O. If U"f is lower semi-continuous for every f then U" « m. If U" « m, then U"f is continuous for every f E btff.

E

btff +,

139

4.1. Spatial Homogeneity

Proof. To prove the first assertion, suppose A E tff and m(A) = O. Then Va(x, A) = 0 for m-a.e. x by (9). Since val A is lower semi-continuous, this implies Va(A) = Va(o,A) = O. Next, if V a « m let ua be a Radon-Nikodym derivative of va with respect to m. We may suppose ua ~ 0 and ua E tff. Then (6) may be written as Vaf(x)

=

f f(x + y)ua(y)m(dy) = f f(z)ua(z -

x)m(dz).

(10)

Since ua E L l(m), a classical result in the Lebesgue theory asserts that

(11) (see e.g. Titchmarsh [1], p. 377). Since shows that Vaf is continuous by (11).

f

is bounded, the last term in (10) D

Proposition 2. The three conditions below are equivalent: (a) (b) (c)

For all a ~ 0 all (X-excessive functions are lower semi-continuous. For some a > 0, all (X-excessive functions are lower semi-continuous. A set A (in tff) is of zero potential if m(A) = O.

Proof. If (b) is true then va « m by Proposition 1. Hence if m(A) = 0, then for every x, m(A - x) = 0 and Va(x, A) = Va(A - x) = O. Thus A is of zero potential. If (c) is true, then va « m. Hence by Proposition 1, Vaf is continuous for every f E bl! +. It follows that Vaf is lower semicontinuous for every f E tff +, by an obvious approximation. Hence (a) is true by Theorem 9 of §3.2, trivially extended to alIa-excessive functions. D

We proceed to discuss a fundamental property of spatially homogeneous Markov processes (X t): the existence of a dual process (XJ The latter is simply defined as follows:

(12)

It is clear that (X t) is a spatially homogeneous process. Let F t , Ga, nt, ft be the quantities associated with it. Thus Ft(x,A) nt(A)

=

= Jrt(-A),

It is convenient to introduce

.

Pl-x, -A) aa(x)

= ua(-x).

4. Brownian Motion

140

Thus we have il'"(x, y)

u'"(y, x)

=

(13)

and U'"(x, A) =

L

(14)

m(dy)u'"(y, x).

(In the last relation it is tempting to write U'"(A, x) for U'"(x, A). This is indeed a good practice in many lengthy formulas.) Now let f E brff +, then U'"f(y) =

f m(dx)f(x)u'"(x, y).

(15)

f u'"(x, Y)J.l(dy)

(16)

Recall the notation V'"J.l(x) =

for a measure J.l. If follows from (15) and (16) that if J.l is a 6-finite measure as well as m, we have by Fubini's theorem:

f U'"f(Y)J.l(dy) = f m(dx)f(x)V'"J.l(x); or perhaps more legibly: (17) This turns out to be a key formula worthy to be put in a general context. We abstract the situation as follows. Let (X t ) and (X t ) be two Hunt processes on the same general (E, rff); and let (Pt), (P t); (V'"), (U'") be the associated semigroups and resolvents. Assume that there is a (reference) measure m and a function u'" ~ 0 such that for every A E rff we have V'"(x, A) =

Lu'"(x, y)m(dy),

U'"(x, A) =

Lm(dy)u'"(y, x).

(18)

Under these conditions we say the processes are in duality. In this case the relation (17) holds for any f E rff + and any 6-finite measure J.l; clearly it also holds with V'" and fr interchanged. Further conditions may be put on the dual potential density function uC,.). We content ourselves with one important result in potential theory which flows from duality. Theorem 3. Assume duality. Let J.l and v be two 6-finite measures such that for some IX ~ 0 we have (19) V'"J.l = V'"v < 00. Then J.l == v.

141

4.1. Spatial Homogeneity

Proof. We have by the resolvent equation for

fJ :;::: a:

Thus if (19) is true it is also true when a is replaced by any greater value. To illustrate the methodology in the simplest situation let us first suppose that J..l and v are finite measures. Take f E bC, then

boundedly because (X t ) is a Hunt process. It follows from this, the duality relation (17), and the finiteness of J..l that

This is also true when J..l is replaced by v, hence by (19) for all large values of a we obtain

This being true for all f

E

bC, we conclude that J..l == v.

In the general case the idea is to find a function h on E such that h > 0 and Jh dJ..l < 00; then apply the argument above to the finite measure h . dJ..l. If J..l is a Radon measure, a bounded continuous h satisfying the conditions above can be easily constructed, but for a general (J-finite J..l we use a more sophisticated method involving an a-excessive function for the dual process (see Blumenthal and Getoor [1]). Since UaJ..l < 00, it is easy to see that there exists g E biS', g > 0, such that by (17):

Put h = Uag; then clearly hE biS', h > 0 and JhdJ..l < 00. Moreover, h being an ()(-potential for (X t ), is a-excessive for (p t ) (called "a-eo-excessive"). Thus it follows from (11) of §3.2 that lim a -+ oo aUah = h. But we need a bit more than this. We know from the Corollary to Theorem 1 of §3.5 that h is finely continuous with respect to (X t ) (called "co-finely continuous"); hence so is fh for any f E bC, because continuity implies fine continuity and continuity in any topology is preserved by multiplication. It follows (see Exercise 7 of §3.5) that lim aUa(fh)

=

fh.

142

4. Brownian Motion

Now we can apply (17) with f replaced by fh to conclude as before that SJhdfl = SJhdv; hence hdfl = hdv; hence fl == v. 0 Next, we shull illustrate the method of Fourier transforms by giving an alternative characterization of a spatially homogeneous and temporally homogeneous Markov process (without reference to sample function regularities). A stochastic process {Xt, t ~ O} is said to have stationary independent increments iff it has the following two properties: (a) (b)

For any 0 ~ to < t i < ... < tn, the random variables {X(t o), X(t k) X(t k - i ), 1 ~ k ~ n} are independent; For any s ~ 0, t ~ 0, X(s + t) - X(t) has the same distribution as X(s) - X(O).

Theorem 4. A spatially homogeneous and temporally homogeneous Markov

process is a process having stationary independent increments, and conversely. Proof. We leave the converse part to the reader (cf. Theorem 9.2.2 of Course). To prove the direct assertion we use Fourier transform (characteristic functions). We write for x E R d, Y E R d : d

<x, y) =

L

(20)

XjYj

j=i

if x

= (Xb ... ,xd),

XU-i)

=

Y = (Yb ... ,Yd)' Let i

=.j=1

and Uk E Rd; and set

o. We have

E{exP[i =

Next we have ifu

E

:t~ X(tk) -

X(tk- i )

JI~~n}

eXP[i kt
R d , s ~ 0, t ~ 0:

E{exp[i
+ t) -

X(t)]IX(t)}

JPs(X(t),dy)exp[i
= exp[ -i
=

Jps(o,dz)exp[i
(22)

143

4.1. Spatial Homogeneity

where ifs is the characteristic function of the probability measure in (2). Taking expectations in (21) we obtain E {ex p

[i

:t:


= E {ex p

Jr s

defined

d)]}

[i kt
X(tk- 1) ]} . if tn + 1 -tJu n+ 1)·

Hence by induction on n, this is equal to n+1

fito(uo)

IT

ntk-tk_,(Uk),

k= 1

where fit is the characteristic function of the distribution fl.t of X(t). Assertion (a) follows from this by a standard result on independence (see Theorem 6.6.1 of Course), and assertion (b) from taking expectations in (22). Note that Jr s is the distribution of X(s) - X(O). D We close this section by mentioning an important property of a process having stationary independent increments: its remote field is trivial. The precise result together with some useful consequences are given below in Exercises 4, 5, and 6. Exercises

1. Is m the unique a-finite measure satisfying (8), apart from a multiplicative constant? [This is a hard question in general. For the Brownian motion semigroup the answer is "yes"; for the Poisson semigroup the answer is "no".]

2. For the uniform motion semigroup (P t ) (Example 2 of §1.2), show that V" « m and find u"(x, y). In this case pix,·) is singular with respect to m for each t and x. For the Poisson semigroup (P t ) (Example 3 of §1.2), find an invariant measure m o (on N). In this case Pt(n,·) « m o for each t and n. 3. The property of independent increments in Theorem 4 may be extended as follows. Let To = 0; for n;::: 1 suppose T n- 1 < Tn and Tn - T n- 1 is optional relative to the a-fields (dt ) , where dr is generated by {X(Tn- 1 + s) - X(Tn-d, 0 < s:s;; t}. Then the random variables X(T n)X(T n- 1 ), n ;::: 1, are independent. The next three exercises are valid for a process having stationary independent increments.

144

4. Brownian Motion

4. Put

where - denotes augmentation as in §2.3. The (J-field 'lJ is called the "remote field of the process. Prove that if A E 'lJ, then r{A} is zero or one for every x. Is the result true for a process having arbitrary independent increments? [Hint: for discrete time analogues see §8.1 of Course.] 5. Suppose there exists t > 0 such that nt has a density. If f E bl! and f = Pt!, then f is a constant. [Hint: show fEe and use the ChoquetDeny-Hunt theorem (Course,§9.5).] 6. Under the hypothesis on (P t ) in Problem 5, a set I! is recurrent or transient (see §3.6) according as PAl == lor PAl =1= 1. [Hint: letf = lim n -+ co t PnPAI, then f is a constant by Exercises 5; now use Problem 4.] 7. Suppose that for each t > 0, there exists a density n; of nt with respect to m such that n; > 0 in E. Then if A E I! and PAl(xo) = 0 for some xo, A must be polar. [Hint: show that PAl(x) = 0 for nt-a.e. x, hence PtPAl(x) = 0 for all x.] 8. Suppose that for each Y, ua (., y) is (X-excessive. (It is possible to polish up ua to achieve this.) Prove that for any measure fl, Uafl is (X-excessive. Under this condition show that the condition (19) in Theorem 3 may be required to hold only m-a.e.

4.2. Preliminary Properties of Brownian Motion The Brownian motion process in R d was introduced in Example 2 of §3.7. For d = 1 it was introduced in Example 4 of §1.2. It is a Markov process with transition function:

Plx, dy) = plx, y) dy, Here dy = dYl ... dYd

is the Lebesgue measure in R d , also written as m(dy). We shall write "a.e." for "m-a.e.", and "density" for "density with respect to m". The product form in (1) shows that the Brownian motion X(t) in R d may be defined as the vector (X1(t), ... ,Xa(t)) where each coordinate Xj(t) is a Brownian motion in R 1, and the d coordinates are stochastically independent. It is spatially homogeneous, and so by Theorem 4 of §4.1 may be characterized as a process having stationary independent increments, such that X(t) - X(O) has the d-dimensional Gaussian (normal) distribution with mean the zero vector

145

4.2. Preliminary Properties of Brownian Motion

and covariance matrix equal to t times the d x d identity matrix. For d = 1, another useful characterization is by way of a Gaussian process, which is defined to be a process having Gaussian distributions for all finite-dimensional (marginal) distributions. The Brownian motion is then characterized as a Gaussian process such that for every sand t: E(X(t))

= 0;

E(X(s)X(t))

=

S 1\

t.

(2)

The proof requires Exercise 2 of §6.6 of Course, which states that mutually orthogonal Gaussian random variables are actually independent. A trivial extension of the definition consists of substituting (J2 t for t in (1), where (J2 > O. We deem this an unnecessary nuisance. In this section we review and adduce a few basic results for the Brownian motion. We begin with the most important one. (I) There is a version of the Brownian motion with continuous sample paths. For d = 1 the proof is given in the Example in §3.1. The general case is an immediate consequence of the coordinate representation (even without independence among the coordinates). This fundamental property is often taken as part of the definition of the process. For simplicity we may suppose all sample functions to be continuous. (II) The Brownian motion semigroup has both the Feller property and the strong Feller property. The latter is verified in §3.7, and the former in Exercise 2 of §2.2. As defined, the strong Feller property does not imply the Feller property! Recall also from Example 4 of §3.7 that Uf is bounded continuous if IE btff and f has compact support, and U~f for I/. > 0 is bounded continuous for f E btff. These properties can often be used as substitutes for the Feller properties. (III) The Brownian motion in Rd is recurrent for d = 1 and d = 2; it is transient for d ::2: 3. This is verified in Examples 2 and 4 of §3.7. Let us mention that for d = 1 and d = 2, there are discrete-time versions of recurrence which supplement the conclusions of Theorem 1 of§3.7. For each 6> 0, let {X(n6), n::2: 1} be a "skeleton" of the Brownian motion {X(t), t::2: O}. Then {X(n6) - X((n - 1)6), n ::2: I} is a sequence of independent and identically distributed random variables, with mean zero and finite second moment (the latter being needed only in R 2 ). Hence the recurrence criteria for a random walk in R 1 or R 2 (§8.3 of Course) yield the following result. For any nonempty open set G: PX{X(n6)

E

G for infinitely many values ofn}

= 1.

This is just one illustration of the close tie between the theories of random walk and of Brownian motion. In fact, all the classical limit theorems have

146

4. Brownian Motion

their analogues for Brownian motion, which are frequently easier to prove because there are ready sharp estimates. Since our emphasis here is on the process we shall not delve into those questions. Let us also recall that for any d 2: 1, each set is either recurrent or transient by Exercise 6 of §4.1. (IV) A singleton is recurrent for Brownian motion in R 1, and polar in Rd for d 2: 2. The statement for R 1 is trivial by recurrence and continuity of paths. The statement for Rd, d 2: 2, is proved in the Example of §3.6. Another proof will be given in §4.4 below. As a consequence, we can show that the fine topology in Rd , d 2: 2, is strictly finer than the Euclidean topology. For example, if Q is the set of points in R d with rational coordinates then R d - Q is a finely open set because Q is a polar set. It is remarkable that all the paths "live" on this set full of holes. (V) Let F = {x E Rdlxl = O},

Gz = {x E Rdlxl < O}.

(3) Then F c G~ 11 G~. To prove this we need consider only the case d = 1, but we will argue in general. Let x E F. By the symmetry of the distribution of Xl (t), we have

PX{X(t)

E

Gd

= PX{X(t) E Gz }.

Since PX{X(t) E F} = 0 by the continuity of the distribution of X1(t), we deduce that PX{X(t) E Gd =!. Hence for each u > 0: PX{X(t) E G1 for all t E [0, u]} ~

t t

which implies Px {T Gz ~ u} 2: !. It follows that Px {T Gz = O} 2: and consequently x E G~ by Blumenthal's zero-or-one law (Theorem 6 of §2.3). Interchanging G1 and Gz we obtain x E G~. Corollary. For the Brownian motion in Rt, each x is regular for {x}.

By (V), the path starting at 0 must enter G1 and Gz in an arbitrarily short time interval [0, e], and therefore must cross F infinitely many times by continuity. Thus 0 is a point of accumulation of the set {o} in the fine topology; the same is true for any x by spatial homogeneity. (VI) For any Borel set B, we have X(TB)EoB

on{O< T B < oo}.

(4)

147

4.2. Preliminary Properties of Brownian Motion

This is a consequence of the continuity of paths, but let us give the details of the argument. By the definition of T B , on the set {TB < oo} for any e > 0 there exists t E [TB , T B + e) such that X(t) E B. Hence either X(T B ) E B, or by the right continuity of the path, X(T B ) = lim X(T B + t) E 13. tHO

On the other hand, on the set {O < T B }, X(t) by the left continuity of the path, X(T B ) = lim X(T B

E

Be for 0 < t < T B , hence

t) E Be.

-

tjjo

Since aB = 13 n Be, (4) is proved. If x E B r , then PX{X(TB ) = X(O) = x} = 1, but of course x need not be in aB, for instance if B is open and x E B. This trivial possibility should be remembered in quick arguments. (VII) Let C n be decreasing closed sets and nn C n = C. Then we have for each x E c e U cr : (5)

This is contained in the Corollary to Theorem 5 of §2A. We begin by the alert that (5) may not be true for x E C - cr! Take for example a sequence of closed balls shrinking to a single point. If x E C, (5) is trivial. Let x E CC. Then there exists k such that x E q. Since q is open x is not regular for Cn for all n 2: k. The rest of the argument is true PX-a.s. We have T Cn > 0 for all n 2: k, and TcJ. Let S = limn i T cn , then 0 < S::s; T c . On {O < T Cn < oo}, we have X(TcJ E C n by (VI). Hence on {O < S < oo}, we have by continuity of paths: X(S) = lim X(TcJ En C n = C. n

n

ThusS2: TcandsoS= Tc.On{S= OO}, T C =

00.

(VIII) Let ~ be either .~? or !F'; (see §2.3). Then for any x, (6)

I1=

are martingales. (Here Ilxl1 2 = 1 xJ.) To verify the second, note that since

IIX(t)11 2 -

d

td =

I

j= 1

(X j (t)2 - t)

148

4. Brownian Motion

we need only verify it for d = 1, which requires a simple computation. Observe that in the definition of a martingale {Xw~~,r}, if the (J-field ff~ is completed to :JF~ with respect to r, then {X t , ff~, r} is also a martingale. Hence we may as well use the completed (J-field. See Exercise 14 below for a useful addition. (IX) For any constant c > 0, (1/c)X(c 2 t), t 2 O} is also a Brownian motion. If we define X(t) = txG)

for t > 0;

=0

for t = 0;

(7)

then {X(t), t 2 O} is also a Brownian motion. The first case is referred to as "scaling". The second involves a kind of reversing the time since lit decreases as t increases, and is useful in transforming limit behavior of the path as t i 00 to t ! O. The proof that X is a Brownian motion is easiest if we first observe that each coordinate is a Gaussian process, and then check in R 1 for s > 0, t > 0:

Thus (2) is true and the characterization mentioned there yields the conclusion. (X) Let B'be a Borel set such that m(B) < such that

00.

sup EX{exp(eTBc)} <

Then there exists e >

00.

°

(8)

xeE

Proof. If x E (13)<, then r{TBc = O} = 1 so (8) is trivial. Hence it is sufficient to consider x E 13. For any t > 0, we have

sup r{TBc > t} :::;; sup PX{X(t) E B} XEB

XEB

f

=

(9)

m(B)

~~~ JB Pt(x, y) dy :::;; (2nt)d/2·

This number may be made :::;;t if t is large enough. Fix such a value of t from here on. It follows from the Markov property that for x E 13 and n 21;

r{TBc > (n

+ l)t} = r{TBc > nt; pX(ntl[TBc > t]} :::;; tr{Tw > nt}

149

4.2. Preliminary Properties of Brownian Motion

since X(nt) E B on {TBc > nt}. Hence by induction the probability above is ::::; 1/2 n + 1. We have therefore by elementary estimation: CXJ

I

W{exp(eTBc)} ::::; 1 +

e,(n+1)tPX{nt < TBc::::; (n + l)t}

n=O

::::; 1 +

CXJ

I

e,(n+1)t2- n.

n=O

This series converges for sufficient small e and (8) is proved. Corollary. If m(B) < PX{TBc < oo} = 1.

00

then

EX{T~c}

<

00

D

for all n 2: 1; in particular

This simple method yields other useful results, see Exercise 11 below. For each r > 0, let us put Tr

= inf{t >

OIIIX(t) -

I

X(O) I 2: r}.

(10)

Let B(x, r) denote the open ball with center x and radius r, namely B(x,r) = {YE

Ellix - yll < r};

(11)

and S(x, r) the boundary sphere of B(x, r): S(x, r) = 8B(x, r) = {y

E

Ellix - yll = r}.

(12)

Then under PX, Tr is just TBe where B = B(x, r); it is also equal to ToB . The next result is a useful consequence ofthe rotational symmetry ofthe Brownian motion. (XI) For each r > 0, the random variables T r and X(Tr) are independent under any P X • Furthermore X(T r ) is uniformly distributed on S(x, r) under P A rigorous proof of this result takes longer than might be expected, but it will be given. To begin with, we identify each OJ with the sample function X( ., OJ), the space Q being the class of all continuous functions in E = Rd. Let cp denote a rotation in E, and CPOJ the point in Q which is the function X(·, CPOJ) obtained by rotating each t-coordinate of X(·, OJ), namely: X

•

X(t, CPOJ) = cpX(t, OJ).

(13)

Since cp preserves distance it is clear that (14)

150

4. Brownian Motion

It follows from (VI) that X(Tr) E S(X(O), r). Let X(O) = x, t 2 0 and A be a Borel subset of S(x, r); put A

= {wi Tr(w)::s:: t; X(Tr(w),w) E A}.

(15)

Then on general grounds we have cp-1A

=

{wi Tr(cpw)::s:: t; X(Tr(cpw), cpw) E A}.

Using (13) and (14), we see that (16) Observe the double usage of cp in cp - 1 A and cp - 1 A above. We now claim that if x is the center of the rotation cp, then (17)

Granting this and substituting from (15) and (16), we obtain

For fixed x and t, if we regard the left member in (18) as a measure in A, then (18) asserts that it is invariant under each rotation cpo It is well known that the unique probability measure having this property is the uniform distribution on S(x, r) given by 0" (A)

O"(r) ,

where

0"

A

E

(19)

g, A c S(x, r)

is the Lebesgue (area) measure on S(x, r); and 2n d/ 2

a(l)

~ r(~)

(20)

Since the total mass of the measure in (18) is equal to P{Tr :s; t}, it follows that the left member of(18) is equal to this number multiplied by the number in (19). This establishes the independence of T r and X(Tr ) since t and A are arbitrary, as well as the asserted distribution of X(Tr). Let us introduce also the following notation for later use: v(r)

r

r

d

= m(B(x,r)) = J~ O"(s)ds = d 0"(1).

(21)

151

4.2. Preliminary Properties of Brownian Motion

It remains to prove (17), which is better treated as a general proposition (suggested by R. Durrett), as follows. Proposition. Let {X t} be a Markov process with transition function (P t). Let cp be a Borel mapping of E into E such that for all t ;:::: 0, X E E and A E Iff, we

have

Then the process {cp(X t)} under px has the same finite-dimensional distributions as the process {X t} under p",(x). Remark. Unless cp is one-to-one, {cp(X t)} is not necessarily Markovian.

Proof. For each f E blff, we have

f Pt(x,dy)f(cpy) f Plcpx,dy)f(y) =

because this true when f = 1A by hypothesis, hence in general by the usual approximation. We now prove by induction on I that for ~ t 1 < ... < t z andfjE blff:

°

The left member of (22) is, by the Markov property and the induction hypothesis, equal to

EX {fl(cpX t,)pX "[f2(cpX t,-t,)· .. !z(cpX tl =

g{fl(cpx t ,)p",x't[f2(Xt2 -

t , ) · ••

tl

,)]}

!z(Xtl -

tl

-,)]}

which is equal to the right member of (22).

D

The proposition implies (17) if A E g;0 or more generally if A

E

g;-.

We conclude this section by discussing some questions of measurability. After the pains we took in §2.4 about augmentation it is only fair to apply the results to see why they are necessary, and worthwhile. (XII) Let B be a (nearly) Borel set, fl and f2 universally measurable, bounded numerical functions on [0, 00) and E a respectively; then the functions (23)

are both universally measurable, namely in lff-jfJ8 where field on R 1 •

f:Ij

is the Borel

152

4. Brownian Motion

That T B E g;-jf14 is a consequence of Theorem 7 of §3.4; next, X(T B ) E g;-j{f follows from the general result in Theorem 10 of §1.3. The assertions of (XII) are then proved by Exercise 3 of §2.4. In particular the functions in (23) are Lebesgue measurable. This will be needed in the following sections. It turns out that if 11 and 12 are Borel measurable, then the functions in (23) are Borel measurable; see Exercise 6 and 7 below. To appreciate the problem of measurability let I be a bounded Lebesgue measurable function from Rd to R 1, and {X t } the Brownian motion in Rd. Can we make I(X t ) measurable in some sense? There exist two bounded Borel functions 11 and 12 such that 11 :s; I :s; 12 and m( {I1 of. I2}) = o. It follows that for any finite measure J.l on {f and t > 0, we have

Thus by definition I(X t ) E /\/1 g;/1 = g;-. Now let T r be as in (10), then under px, X(T r) E S(x, r) by (VI). But the Lebesgue measurability of I does not guarantee its measurability with respect to the area measure (J on S(x, r), when I is restricted to S(x, r). In particular if 0 :s; I :s; 1 we can alter the 11 and 12 above to make 11 = 0 and 12 = 1 on S(x, r), so that E"{I2(X t ) f1 (X t )} = a(S(x, r)). It should now be clear how the universal measurability of I is needed to overcome the difficulty in dealing with f(X(T r )). Let us observe also that for a general Borel set B the "surface" aB need not have an area. Yet X(T oB ) induces a measure on aB under each P/1, and if I is universally measurable £i1{I(X(ToB ))} may be defined. Exercises

Unless otherwise stated, the process discussed in the problems is the Brownian motion in R d, and (Pt) is its semigroup. 1. If d 2: 2, each point x has uncountably many fine neighborhoods none of which is contained in another. For d = 1, the fine topology coincides

with the Euclidean topology. 2. For d = 2, each line segment is a recurrent set; and each point on it is regular for it. For d = 3, each line is a polar set, whereas each point of a nonempty open set on a plane is regular for the set. [Hint: we can change coordinates to make a given line a coordinate axis.] 3. Let D be an open set and XED. Then PX{TDc = T oD } = 1. Give an example where r {TDC < TO))c} > O. Prove that if at each point y on aD there exists a line segment yy' E DC, then PX{TDc = T(D)e} = 1 for every xED. [Hint: use (V) after a change of coordinates.] 4. Let B be a ball. Compute EX {TBe} for all x E E. 5. If BE {f and t > 0, then r{T B = t} = O. [Hint: show Sr{TB = s} dx = 0 for all but a countable set of s.]

153

4.2. Preliminary Properties of Brownian Motion

6. For any Hunt process with continuous paths, and any closed set C, we have T c E ;yo. Consequently x ----> PX{Tc:S; t} and x ----> EX{j(X(Tc)) are in ~ for each t 2: and I E b~ or ~ +. Extend this to any C E ~ by Proposition 10 of §3.5. [Hint: let C be closed and Gn tt C where Gn is open; then nn{:3tE[a,b]:X(t)EGn} = {3tE[a,b]:X(t)EC}.] 7. For any Hunt process satisfying Hypothesis (L), B E~, IE f!d and I E~ + respectively, the functions x ----> EX{j(TB )} and x ----> EX{j(X(TB ))} are both in ~. [Hint: by Exercise 3 of §3.5, x ----> EX{e-"TB} is in ~ for each il( > 0; use the Stone-Weierstrass theorem to approximate any function in Co([O, (0)) by polynomials of e- x . For the second function (say ({J) we have il( U"({J ----> ({J if I E bC +. This is due to Getoor.]

°

8. If IE L l(R d ) and t > 0, then Ptf is bounded continuous. 9. If I E ~ and ptlll < 00 for every t > 0, then Ptf is continuous for every t > 0. [Hint: for Ilxll:s; 1 we have Iptf(x) - Ptf(o) I:s; A[Ptlll(o) + P4tlll(0)]. This is due to T. Liggett.] 10. If IE M, then limt~oo [Ptf(x) - Ptf(y)] = put x- z= J2i.( in Jlplx, z) - Pt(y, z)1 dz.]

°

for every x and y. [Hint:

11. In (X) if e is fixed and m(B) ----> 0, then the quantity in (8) converges to one.

°

12. Let X be the Brownian motion in R 1, a > and B be a Borel set contained in (- oo,a]. Prove that for each t > 0: pO{T{al:S; t; X(t) E B} = pO{X(t) E 2a - B} and deduce that pO{T{al:S; t} = 2pO{X(t) > a}. This is Andre's reflection principle. A rigorous proof may be based on Theorem 3 of §2.3. 13. For Brownian motion in R 1, we define the "last exit from zero before time t" as follows: y(t) = sup{s :s; t:X(s) = O}. Prove that for s E (0, t) and x E

pO{y(t)

E

R\ we have

PO{y(t)

E

ds} =

ds; X(t)

E

dx} =

ds 1i~s(t-s)

,

x

21i~s(t -

s?

e- x2 / 2 (t-S) ds dx.

[This gives a glimpse of the theory of excursions initiated by P. Levy. See Chung [8] for many such explicit formulas. The literature is growing in this area.] 14. For each
AE R

d

,

{exp(A,X t )

= I~=l AjXj.

-IIAI1 2 t/2,!Fr, PX}

is a martingale, where

154

4. Brownian Motion

4.3. Harmonic Function A domain in R d ( = E) is an open and connected (nonempty!) set. Any open set in R d is the union of at most a countable number of disjoint domains, each of which called a component. It will soon be apparent that there is some advantage in considering a domain instead of an open set. Let D be a domain. We denote the hitting time of its complement by 'D' namely: 'D = TDc = inf{t > 0IX(t)

E

DC}.

(1)

This is called the "(first) exit time" from D. The usual convention that'D = + 00 when X(t) E D for all t > is meaningful; but note that X( (0) is generally not defined. By (X) of§4.2, we have PX{'D < oo} = 1 for all XED ifm(D) < 00, in particular if D is bounded. The class of Borel measurable functions from R d to [ - 00, + 00] or [0, 00] will be denoted by g or g +; with the prefix "b" when it is bounded. We say f is locally integrable in D and write f E Lloc(D) iff SK If Idm < 00 for each compact KeD, where m is the Lebesgue measure. We say A is "strictly contained" in B and write A @ B iff A c B. Let f E g +, and define a function h by

°

h(x) = W{j(X('D));

'D <

oo}.

(2)

This has been denoted before by P'Df or PDcf. Although h is defined for all x E E, and is clearly equal to f(x) if x E (DC)O = (15)", we are mainly concerned with it in 15. It is important to note that h is universally measurable, hence Lebesgue measurable by (XII) of§4.2. More specifically, it is Borel measurable by Exercise 6 of §4.2 because DC is closed and the paths are continuous. Finally if f is defined (and Borel or universally measurable) only on aD and we replace 'D by TaD in (2), the resulting function agrees with h in D (Exercise 1).

Recall the notation (11), (12) and (20) from §4.2. Theorem 1. If h

1=

00

in a domain D, then h <

00

in D. For any ball B(x, r)@D,

we have h(x) =

~( )i a r JS(x,r)

h(y)a(dy).

(3)

Furthermore, h is continuous in D. Proof. Write B for B(x, r), then almost surely 'B < 00; under p x , 'B < 'D so that'D = 'B + 'D 8'B' Hence it follows from the fundamental balayage formula (Proposition 1 of §3.4) that 0

(4)

155

4.3. Harmonic Function

which is just (5)

This in turn becomes (3) by the second assertion in (XI) of §4.2. Let us observe that (4) and (5) are valid even when both members of the equations are equal to + 00. This is seen by first replacing f by f 1\ n and then letting n ~ 00 there. A similar remark applies below. Replacing r by s in (3), then multiplying by a(s) and integrating we obtain h(x) jr a(s) ds

Jo

=

j h(y)a(dy) ds Jojr JS(x,s)

=

j

JB(x,r)

h(y)m(dy).

(6)

where in the last step a trivial use of polar coordinates is involved. Recalling (21) of §4.2 we see that (6) may be written as h(x) = _(1) j

vr

JB(x,r)

h(y)m(dy),

(7)

whether h(x) is finite or not. Now suppose h =1= 00 in D. Let h(x o) < 00, X o E D; and B(x, p) c B(x o, r) D. It follows from (7) used twice for x and for X o that h(x)

= _(1) j v p

JB(x,p)

h(y)m(dy)

@

~ _(1) j h(y)m(dy) v p JB(xo,r) v(r) ~ v(p) h(xo) <

00.

(8)

Thus h < 00 in B(xo,r). This shows that the set F = D n {x Ih(x) < oo} is open. But it also shows that F is closed in D. For if X n E F and X n ~ X oo ED, then X oo E B(x n , <5) @ D for some <5 > and some large n, so that X oo E F by the argument above. Since D is connected and our hypothesis is that F is not empty, we conclude that F = D. The relation (7) then holds for every x in D with the left member finite, showing that h E Lloc(D). To prove that h is continuous in D, let r be so small that both B(x, r) and B(x', r) are strictly contained in D. Applying (7) for x and x' we obtain

°

Ih(x) - h(x')1

~ v~r) Ie h(y)m(dy)

(9)

where C = B(x, r) D. B(x', r). If x is fixed and x' ~ x then it is obvious that m( C) ---> 0, and the integral in (8) converges to zero by the proven integrability of h. Thus h is continuous and Theorem 1 is proved. D We have shown above that the "sphere-averaging" property given in (3) entails the "ball-averaging" property given in (7). The converse is also true.

156

4. Brownian Motion

For we may suppose h"¥= 00 in D; then h is finite continuous in D by the proof above. Now write (7) as follows: h(x)v(r) =

rr r

Jo JS(x,sl

h(y)O'(dy) ds.

Since h is continuous its surface integral over Sex, s) is also continuous in s. Differentiating the equation above with respect to I' we obtain (3) since v'(r) = 0'(1'). We now proceed to unravel the deeper properties of the function h. Definition 1. Let D be an open set. A function h is called harmonic in D iff

(a) (b)

it is finite continuous in D; the sphere-averaging property holds for any B(x, r)

§

D.

By the proof of Theorem 1, if D is a domain, and h is locally integrable in D, then (b) entails (a). Another easy consequence is as follows. Corollary. Let D be a domain and f be a Borel measurable function on aD. If Pvclfl "¥= 00 in D, then Pvc! is harmonic in D. In particular this is the case iff is bounded on aD.

It is remarkable that there is another quite different characterization of the class of harmonic functions. Recall that C(kl(D) is the class of k-times continuously differentiable functions in D; Ll

=

(8)2

L -8x d

j=l

j

is the Laplacian in Rd. Definition 2. A function h is harmonic in the open set D iff it belongs to C(2l(D), and satisfies in D the Laplace equation:

Llh = O.

(10)

Theorem 2. (Gauss-Koebe) Definitions 1 and 2 are equivalent. Moreover a harmonic function in D belongs to C(OOl(D).

Proof. Let h be harmonic in D according to Definition 1. We show first that hE C(OOl(D). For any c5 > 0 there exists a function cp in C(OOl(E) with the following properties: cp(x) = cp(llxli):cp(x) > 0 for Ilxll < c5; cp(x) = 0 for Ilxll :?: c5; and h cp(x)m(dx) = 1. See Exercise 6 for an example. We have then

foOO cp(r)O'{r) dr

=

1.

(11)

157

4.3. Harmonic Function

For each x in D, there exists £5 > 0 such that h is bounded in B(x, (5) and (3) holds for 0 < r < <5. It follows that hex)

CD

JI0

=

= Jo ,CD

[~( )f 0' r

S(x,r)

f

S(x,r)

h(Y)O'(dY)]
h(y)
(12)

= IE h(y)
JB

= , aah (y)O'(dy) JoB n

(13)

where ahlan is the outward normal derivative. Now let us denote the right member of (3) by A(x, r). Making the change of variables y = x + rz we have A(x, r)

= 0' (11)' hex + rz)O'(dz). JS(o,l)

(14)

Straightforward differentiation with respect to r gives for fixed x: I

hex

d

+ rz) = '~1

Zj

ah az. (x

J-

I

A (x, r)

=

(15)

J

f - _1_ f - O'(r)

1 = ~(1) 0'

ah

+ rz) = an (x + rz);

S(o,l)

ah -an (x

S(x,r)

ah ( )O'(d ) an y y

~( )' 0' r

JB(x,r)

+ rz)O'(dz)

Llh(y)m(dy).

(16)

The differentiation of(13) under the integral is correct because h has bounded continuous partial derivatives in a neighborhood of Sea, 1). The first member of (16) vanishes for all sufficiently small values of r by (3), hence so does the integral in the last member (rid of the factor 1/00(r)). This being true for all

158

4. Brownian Motion

sufficiently small values of r, we conclude Llh(x) = 0 by continuity. Thus h is harmonic according to Definition 2. Conversely if that is the hypothesis, then for any B(x, r) @ D, the formula (16) is valid because hE e(2)(D). Since Llh = 0 in D the last member of(16) vanishes, hence so does the first member. Hence A(x, r) = lim dO A(x, r), which is equal to h(x) by the continuity of h. Therefore (3) is true and Theorem 2 is proved. D Definition 2 allows a "local" characterization ofharmonicity. The function h is harmonic at x iff it belongs to e(2) in a neighborhood of x and Llh(x) = O. It also yields an improvement of Definition 1, by retaining condition (a) there

but weakening condition (b) as follows: (b') the sphere-averaging property holds for each x in D, and all sufficiently small balls (strictly) contained in D with center at x. Actually, even weaker conditions than (b') suffice, but these are technical problems; see Rao [1]. We add two basic propositions about harmonic functions below, both deducible from Definition 1 alone. The first is sometimes referred to as a "principle of maximum." However, there are some other results in potential theory called by that name. Proposition 3. Let h be harmonic in the domain D, and put

M = sup h(x), XED

m

=

inf h(x). XED

If there exists an x in D such that h(x) = M or h(x) = m, then h is a constant in D. Proof. Suppose h(x) = M. Then it follows at once from the ball-averaging property (7) that h(y) = M first for a.e. y in a neighborhood of x, then for all y there since h is continuous. Thus the set D n {x I h(x) = M} is open in D; it is also relatively closed in D by continuity. Hence h = M in D. The proof for the infimum is exactly the same. D Corollary. If h is harmonic in a bounded domain D, and continuous in D, then the maximum and minimum values of h in D are taken on aD. In particular if h == 0 on aD, then h == 0 in D.

The next proposition is one of Harnack's theorems which hark back to Theorem 1. We relegate two other Harnack's theorems to Exercises 8 and 9. Proposition 4. Let D be a domain and {h m n ~ I} a sequence of harmonic functions in D. Suppose hn increases to hoo in D. Then either h oo == + 00 in D, or h is harmonic in D. Proof. Even without the use of Definition 2, it is easy to see that we need only prove that h oo is harmonic in each bounded subdomain D 1 @ D. On Db

159

4.3. Harmonic Function

h 1 ;;::: m > - 00 by continuity. Let hn = hn - m; then hn increases to h oo = h oo - m ;;::: O. Each hn has the sphere-averaging property in Db hence so does hoo by monotone convergence. Therefore by the proof of Theorem 1 from (6) on, either hoo == + 00 in Db or hoo is finite continuous, hence harmonic inD 1 · D Thanks to Definition 2, it is trivial to find explicit harmonic functions. It is well known that the real part of a complex analytic function is harmonic (in R Z ), see e.g. Ahlfors [1]. Indeed the two characterizations have their analogues in the Cauchy integral formula and holomorphy, respectively. It is obvious from Definition 2 that in R 1, harmonicity is tantamount to linearity. EXAMPLE. In R Z , logllxll is harmonic in any open set not containing o. In Rd , d ;;::: 3, 1/llxlld-z is harmonic in any open set not containing o. Let K be a compact set, J.l a measure such that J.l(K) < 00; then (17) are harmonic functions of x in R Z - K and R d - K respectively. The first two assertions are standard exercises in calculus upon using Definition 2. The last assertion then follows either by differentiation under the integrals, or perhaps better by integration using Definition 1 and Fubini's theorem. Now consider the particular case of (17) when J.l is the uniform distribution on the sphere S(o, r). If Ilxll > r, then z ~ logllzll is harmonic in a neighborhood of B(x, r); hence the sphere-averaging property yields logllxll = _(1) a r

r logllx - Ylla(dy); JrS(x,r) logllzlla(dz) = _(1) a r JS(o,r)

similarly for the second integral in (17). However, to evaluate the integrals when Ilxll < r we need the following proposition. Proposition 5. If h is harmonic and is a junction of Ilxll alone, then it must be of the form

(18)

respectively in R Z and R d , d;;::: 3: where

C1

and C z are constants.

Proof. Writing p for Ilxll, we see that the Laplace equation for h reduces to dZ d- 1 d dpz h + -p- dp h = O.

(19)

160

4. Brownian Motion

Solving this simple differential equation we obtain the general solutions given in (18). These are called fundamental radial solutions of the Laplace equation. D Proposition 6. We have for any r > 0:

~) O"(r Jr

S(o,r)

logllx - YIIO"(dy) = 10g(llxll v r),

~) r Ilx O"(r JS(o,r)

YIIZ-dO"(dy) = (11xll v r)Z-d,

d =2;

(20)

d 2 3.

(21)

Proof. Denote the left member of (20) by f(x). It is harmonic in B(o, r) and is a function of Ilxll alone upon a geometric inspection. Hence by Proposition 5, f(x) = a logllxll + b. For x = 0 we have f(o) = log r by inspection; hence a = 0 and b = log r. It follows that f(x) = log r for Ilxll < r. Similarly, f(x) = a' logllxll + b' for Ilxll > r. It is easily seen that limllxll-+<XJ [f(x) -logllxllJ= 0; hence a' = 1 and b' = O. Thus f(x) = logllxll for Ilxll > r. We have already observed this above. It remains to show thatf is finite continuous at Ilxll = r. This is a good exercise in calculus, not quite trivial and left to the reader. The evaluation of (21) is similar. D

The exact calculation of formulas like (20) and (21) forms a vital part of classical potential theory, and is the source of many interesting results. Exercises 1. For any open set D show that PX{ X(-rD)

E aD} = 1 for all x E 1} However, if x E aD it is possible that r{r D < TaD} = 1. An example is known as Lebesgue's thorn, see Port and Stone [1], p. 68.

2. Show that if h is harmonic then so are all its partial derivatives of all orders. Verify the harmonicity of the following functions in R 3 : Xl

Ilxll '

R1

tan

-1 X z Xl'

x 3 tan

-1 X z Xl'

Z 2X 3 -

Z Xl -

Z Xz·

3. Let h be harmonic in R d, d 2 1 such that Ptlhl < ctJ for all t > O. Then h = Pth for all t 2 0, where (P t) is the semigroup of the Brownian motion. The process {h(X t ), g;;, t 2 O} is a continuous martingale. [Hint: use polar coordinates and the sphere-averaging property; a by-product is the value of 0"(1).] 4. Let h E iff, 0 :::;; h < 00 and h = P 1 h. Then h is a constant. [Hint: if h =i= ctJ then h is locally integrable; now compare the values of Pnh at two points as n -+ ctJ, by elementary analysis. This solution is due to Hsu Pei.]

161

4.3. Harmonic Function

5. (Picard's Theorem.) If h is harmonic and :::::-:0 in R d , then it is a constant. [Hint: there is a short proof by ball-averaging. A longer proof uses martingale and Exercise 4 of §4.1 to show that lim t _ oo h(Xt ) is a constant a.s.] 6. Let <,D1(X) = c 1 exp((llxl12 - 1)-1) for Ilxll < 1, <,D1(X) = 0 for Ilxll : : :-: 1, and d C 1 is so chosen that SRd <,D1(x)dx = 1. Put <,Db(X)=<,D1(X/b)(I/b ). Show that <,Db has all the properties required of <,D in the proof of Theorem 2. 7. Let D be a domain and h harmonic in D. If D is unbounded we adjoin a point at infinity 00 to aD, and say that x -+ 00 when Ilxll-+ 00. Suppose now that lim

h(x)

s

M.

(22)

D3X-aDu{oo}

Then h(x) s M for all XED. Give an example in which the conclusion becomes false if { OCJ } is omitted in (22). 8. Let {h,,} be a family of harmonic functions in an open set D. Suppose that the family is uniformly bounded on each compact subset of D. Then from any sequence from the family we can extract a subsequence which converges to a harmonic function in D, and the convergence is uniform on each compact subset of D. [Hint: prove first equi-continuity of the family on each compact subset by using (9), then apply the Ascoli-Arzela theorem.] 9. Let D be a domain. For each compact KeD there exists a constant c(K) such that for any harmonic function h > 0 in D we have for any Xl E K, X2 EK: h(x 1 ) h(x ) 2

:::::-:

c(K).

[Hint: the classical proof uses inequalities which are derived from the Poisson representation for a harmonic function in a ball (see §4.4). But a proof follows from (8), followed by the usual "chain argument."]. 10. Let a > 0, b > O. Evaluate the following integrals by means of Proposition 6: S02" log(a 2 + b2 - 2ab cos B)dB;

S:" (a

2

+ b2

-

2ab cos B) -1/2 dB.

Compare the method with contour integration in complex variables. 11. Prove the continuity of the left members of (20) and (21) as functions of

x, as asserted in Proposition 6.

4. Brownian Motion

162 12. Compute

r

JB(o,r)

r

logllx - Yllm(dy),

JB(o,r)

Ilx -

YI12-d m (dy)

in R 2 and R d , d ~ 3, respectively. [Answer for the second integral: a(l)

d

rd Ilxlld-2

r2 a(i

if Ilxll ~ r;

+ a(I)ll x Il 2

G-~).

if Ilxll ::; r.]

13. (Gauss) In R d , d ~ 3, let ~ eKe B(o, r), then ~-2

j1(K)

= - ()

ad r

r

JS(o,r)

U j1(y)a(dy).

Here aAr) = a(r) in (20) of §4.2. 14. Extend the result in Exercise 13 to R 2 , using U* (given in (14) of §4.6 below) for U.

4.4. Dirichlet Problem Let D be an open set and f E b~ on aD, and consider the function h defined in 15 by (2) of §4.3. Changing the notation, we shall denote h by H nf, thus X E

D.

(1)

This formula defines a measure H n(x, .) on the boundary aD, called the harmonic measure (of x with respect to D). For each Borel subset A of aD, the function H(', A) is harmomic in D by the Corollary to Theorem 1 of §4.3. We may regard the measure H(x,') as defined for all Borel sets, though it is supported by aD. The measure H(x,') is also defined for x E DC and is harmonic in (15)' = (DT For example if aD is the horizontal axis in R 2 , H Df is a harmonic function both in the upper and lower open half-planes, but not necessarily in the whole plane. We now study the behavior of h(x) as x approaches the boundary. Since h is defined only in 15, we shall not repeat the restriction that XED below. Recall that a point z on aD is regular or irregular for DC according as pz {T D = O} = 1 or 0 by Blumenthal's zero-or-one law. It turns out that this is the same as its being a "regular boundary point of D" in the language of the non-probablistic treatment of the Dirichlet problem. Since z may be regular for DC but irregular for D, the difference of terminology should be borne in mind. For instance, it is easy to construct a domain with a boundary point z which is irregular for DC, but not so easy for z to be irregular for D.

163

4.4. Dirichlet Problem

Proposition 1. For each t > 0, the function

is lower semi-continuous in E. Proof. Since D is open and X is continuous, we have {T D = t} It follows that {T D> t} = {\is E (O,t]:X(s) E D}

C

{X(t)

E

DC}.

nOl {\iSEG, t}X(S)ED}. The set above is in

ff?, and (2)

The probability on the right side of (2) is of the form pt/n
= t} = 0, but we do not

Theorem 2. If z on aD is regular for DC, and if f is continuous at z, then we have

lim h(x)

=

(3)

f(z).

D3X---I-Z

Proof. We have by Proposition 1, for each 8> 0:

lim PX{T D ::;; 8} ;::: PZ{TD::;; 8}

= 1.

(4)

x-->z

Hence the corresponding limit exists and is equal to one. Note that here x need not be restricted to 15. Let T r be as in (10) of §4.2. For r > and 8> 0, PX{Tr > 8} does not depend on x, and for any fixed r we have

°

lim PO{Tr > 8}

= 1,

(5)

etO

because T r >

°

almost surely. Now for any x, an elementary inequality gives

164

4. Brownian Motion

Consequently we have by (4) (7) X~Z

and so by (5):

(8)

On the set {'v < T r }, we have IIX(,v) - X(O)II < r and so IIX(,v) - zll < IIX(O) - zll + r. Since f is continuous at z, given b > 0, there exists 1] such that if y E aD and Ily - zll < 21], then If(Y) - f(z)1 < b. Now let xE 15 and Ilx - zll < 1]; put r = 1]. Then under PX, X('v) E aD (Exercise 1 of §4.3) and IIX(,v) - zll < 21] on {'v < T~}; hence If(X(,v)) - f(z) 1< 15. Thus we have

EX{lf(X(,v)) - f(z) 1; 'v < oo} < <5PX{'v < T~}

+ 21Ifllpx{T~ ~ 'v < oo}.

(9)

When x ....... z, the last term above converges to zero by (8) with 1] for r. Since 15 is arbitrary, we have proved that the left member of (9) converges to zero. As a consequence, lim h(x) = f(z) lim PX{'v < CIJ} = f(z) X~Z

by (4).

D

Let us supplement this result at once by showing that the condition of regularity of z cannot be omitted for the validity of (3) in R d , d 2 2. Note: in R 1 every z E aD is regular for {z} cDC. Theorem 3. Suppose d 2 2 and z is irregular for DC. Then there exists f E be on aD such that (10) lim h(x) < f(z). D~z

Proof. Indeed (10) is true for any f E bqaD) such that f(z) = 1 and f < 1 on {z}. An example of such a function is given by (1 -llx - zl!)vO. Since {z} is polar by (IV) of §4.2, and z is irregular for DC, we have P X('v) = z} = O. This implies the first inequality below: aD -

Z

{

1 > EZ{j(X(,v)); 'v < oo}

= lim EZ{Tr < 'v; h(X(Tr))} dO

2 lim PZ{Tr < 'v} rtO

inf

h(x).

xEB(z,r)nV

where the equation follows from limdO PZ{Tr < 'v} = 1, and the strong Markov property applied at T r on {T r < 'v}. The last member above is equal to the left member of (10). D

165

4.4. Dirichlet Problem

We proceed to give a sufficient condition for z to be regular for DC. This is a sharper form of a criterion known as Zaremba's "cone condition". We shall replace his solid cone by a flat one. Let us first observe that if z E aD and B is any ball with z as center, then z is regular for DC if and only if it is regular for (D (\ BY. In this sense regularity is a local property. A "flat cone" in R d , d ~ 2, is a cone in a hyperplane of dimension d - 1. It is "truncated" if all but a portion near its vertex has been cut off by a hyperplane perpendicular to its axis. Thus in R 2 , a truncated cone reduces to a (short) line segment; in R 3 , to a (narrow) fan.

Theorem 4. The boundary point z is regular for DC cone with vertex at z and lying entirely in DC.

if there

is a truncated flat

Proof. We may suppose that z is the origin and the flat cone lies in the hyperplane of the first d - 1 coordinate variables. In the following argument all paths issue from o. Let Tn

= inf{t >

~IXAt) = o}.

Since 0 is regular for {o} in R 1 by (V) of §4.2, we have limn~ 00 Tn = 0, PO-a.s. Since XA') is independent of Y(') = {X 1('), ... ,X d - 1 (')}, Tn is independent of Y(·). It follows (proof?) that Y(T n ) has a (d - l}-dimensional distribution which is seen to be rotationally symmetric. Let C be a flat cone with vertex 0, and Co its truncation which lies in DC. Then rotational symmetry implies that po {Y(Tn) E C} = 8, where 8 is the ratio of the "angle" of the flat cone to a "full angle". Since limn~oo Y(Tn ) = 0 by continuity, it follows that limn~ 00 po {Y(T n) E Co} = 8. Since Co c DC, this implies

n~oo

Therefore,

0

is regular for DC by the zero-or-one law.

o

Let us call a domain D regular when every point on aD is regular for DC. As examples of Theorem 4, in R 2 an open disk with a radius deleted is regular; similarly in R 3 an open ball minus a sector of its intersection with a plane

166

4. Brownian Motion

is regular, see Figure (p. 165). Of course, a ball or a cube is regular; for a simpler proof, see Exercise 1. The most classical form of the Dirichlet boundary value problem may be stated as follows. Given an open set D and a bounded continuous function f on aD, to find a function which is harmonic in D, continuous in 15, and equal to f on aD. We shall refer to this problem as (D,f). The Corollary to Theorem 1 of §4.3, and Theorem 2 above taken together establish that the function Hvf is a solution to the problem, provided that D is regular as just defined. When B is a ball, for instance, this is called the "interior Dirichlet problem" if D = B, and the "exterior Dirichlet problem" if D = (BY. The necessity of regularity was actually discovered long after the problem was posed. If there are irregular boundary points, the problem may have no solution. The simplest example is as follows. Let D = B(o, 1) - {o} in R 2 , namely the punctured unit disk; and define f = 0 on aB(o, 1), f(o) = 1. The function H vf reduces in this case to the constant zero (why?), which does not satisfy the boundary condition at o. But is there a true solution to the Dirichlet problem? Call this hI; then hI is harmonic in D, continuous in 15 and equal to and 1 respectively on aB(o, 1) and at o. Let cp be an arbitrary rotation about the origin, and put h2 (x) = hl(cpx) for x E D. Whether by Definition 1 or 2 of §4.3, it is easy to see that h 2 is harmonic in D and satisfies the same boundary condition as hI' Hence hI is identical with h 2 (see Proposition 5 below). In other words, we have shown that hI is rotationally invariant. Therefore by Proposition 5 of §4.3, it must be of the form C I logllxll + C 2 in D. The boundary condition implies that C I = 0, C2 = 1. Thus hI is identically equal to one in D. Ii. cannot converge to zero at the origin. [The preceding argument is due to Ruth Williams.] It is obvious that 0 is an irregular boundary point of D, but one may quibble about its being isolated so that the continuity of f on aD is a fiction. However, we shall soon discuss a more general kind of example by the methods of probability to show where the trouble lies. Let us state first two basic results in the simplest case, the first of which was already used in the above.

°

Proposition 5. If D is bounded open and regular, then H vf is the unique solution to the original Dirichlet problem (D,f). Proof. We know that h = Hvf is harmonic in D. Since all points of aD are regular, and f is continuous in aD, h is continuous in 15 by Theorem 2. Let hI be harmonic in D, continuous in 15, and equal to f on aD. Then h - hI is harmonic in D, continuous in 15, and equal to zero on aD. Applying the D Corollary to Proposition 3 of §4.3, we obtain h - hI = in D.

°

The following result is to be carefully distinguished from Proposition 5. Proposition 6. Let D be a bounded open set. Suppose that h is harmonic in D and continuous in 15, Then h(x)

=

H vh(x),

\Ix

E

D.

167

4.4. Dirichlet Problem

Proof. There exist regular open sets Dn such that Dn it D. In fact each D n may be taken to be the union of cells of a grid on R d , seen to be regular by the cone condition. We have h = HDnh in 15n by Proposition 5. Let first xED, then there exists m such that x E Dn for n ~ m; hence

Letting n ~ 00, we have T Dn i T D PX-a.s. by (VII) of §4.2. Hence the right member above converges to H Dh(x) by bounded convergence. Next if x E aD and x is regular for DC, then HDh(x) = h(x) trivially. Finally if x E aD and x is not regular for DC, then PX{TD > O} = 1. We have for each t > 0: EX{t <

TD ;

h(X(TD))} = EX{t <

TD ;

EX(tl[h(X(T D))]}

<

TD ;

h(X(t))}

= EX{t

because X(t) E Don {t < T D}, so the second equation follows from what we have already proved. As t 1 0, the first term above converges to H Dh(x), the third term to EX{O < TD; h(X(O))} = h(x) by bounded convergence. D We are now ready to discuss a general kind of unsolvable Dirichlet problem. Let Do be a bounded open domain in Rd , d ~ 2, and S a compact polar set contained in Do. For instance, if d = 2, S may be any countable compact set; if d = 3, S may be a countable collection oflinear segments. Let D = Do - S, and define f = Ion aDo, f = 0 on S. Note that aD = (aDo) U S since a polar set cannot have interior points. There exists a sequence of open neighborhoods B n 11 S such that En C Do. Let Dn = Do - B.. Suppose that h is harmonic in D, continuous in 15, and equal to one on aDo. Nothing is said about its value on S. Since h is harmonic in Dn and continuous in 15., we have by Proposition 6 for x E Dn : h(x) = EX{h(X(TDJ)} r{TODo < ToBJ

=

+ W{TOBn <

T oDo ; h(X(ToBJ)}.

(11)

Since S is polar, we have lim r{TOBn < T oDo } = r{Ts < T oDo } = O.

(12)

n

Since h is bounded in 15, it follows that the last term in (11) converges to zero as n ~ 00. Consequently we obtain h(x)

=

lim r{TODo < ToBJ n

=

r{TODo < oo}

=

1.

(13)

This being true for x in Dn for every n, we have proved that h == 1 in D. Thus the Dirichlet problem (D,f) has no solution.

168

4. Brownian Motion

More generally, let f be continuous on aDo, and arbitrary on S. Then H Df is harmonic in D, and the argument above shows that H Do! is the unique continuous extension of H Df to Do (which is harmonic in Do to boot). In particular, the Dirichlet problem (D,f) is solvable only if f = H Do! on S. In the example above we constructed an irregular part of aD by using the polar set S. We shall see in §4.5 below that for any open D, the set of irregular points of aD is always a polar set. This deep result suggests a reformulation of the Dirichlet problem. But let us consider some simple examples first.

1. In R d , d 2:: 3, let B = B(o, b). Consider the function PB 1(x) for E Be. This function is harmonic in Be and is obviously a function of alone. Hence by Proposition 5 of §4.3, it is of the form + Cz. The following equation illustrates a useful technique and should be argued out directly:

EXAMPLE

x

clllxllz-d

Vx

E

Be: U(x, B) = PX{TB < IX)} U(y, B)

Ilxll

(14)

where y is any fixed point on aBo But actually this is just the identity U1 B = P B U1 B contained in Theorem 3(b) of §3.4, followed by the observation that U(y, B) is a constant for y E aB by rotational symmetry. Now the left member of (14), apart from a constant, is equal to

r

dy

JB Ilx _ ylld-Z which converges to zero as IIXII-4 00. Hence P B 1(x) -4 0 as d Z z = O. Since P B 1 = 1 on aB, C 1 = b - • Thus .

IIXII-4 00, and so

C

r{TB
00

b )d-Z ' (W

Ilxll 2:: b.

(15)

and

D =

{xla <

Ilxll < b}.

(16)

Consider the Dirichlet problem (D,f), where f is equal to one on S(o, a), zero on S(o, b). The probabilistic solution given by PDcf signifies the probability that the path hits S(o, a) before S(o, b). It is the unique solution by Proposition 5. Since it is a function of Ilxll alone it is of the form C-LIIXllz-d + C z . The constants C 1 and C z are easily determined by its value for Ilxll = a and Ilxll = b. The result is recorded below: a ~

Ilxll ~ b

(17)

169

4.4. Dirichlet Problem

If b i + 00 then TS(o,b) i 00 (why?), and we get (15) again with b replaced bya. Exactly the same method yields for d = 2; x _ logllxll - log b P {TS(o,a) < TS(o,b)} - I og a - I og b'

a::; Ilxll ::; b.

(18)

If we let b i + 00, this time the limit ofthe above probability is 1, as it should be on account of recurrence. On the other hand, if we fix b and let a lOin (18), the result is PX{T{o} < TS(o,b)} = 0 (why?). Now let biro to conclude that P{o}l(x) = 0 for x =1= o. Since Pix, {o}) = 0 for any x in E and t > 0, it follows that P t P{o}l(x) = 0; hence P{o}l(x) = lim qo P t P{o}l(x) = O. Thus {o} is a polar set. This is the second proof of a fundamental result reviewed in (IV) of §4.2.

EXAMPLE 2. The harmonic measure H B(X, .) for a ball B = B( 0, r) can be explicitly computed. It is known as Poisson's formula and is one of the key formulas in the analytical developments of potential theory. In R 2 it is an analogue of, and can be derived from, the Cauchy integral formula for an analytic function in B:

f(x)

f

1 -fez) -dz 2ni S(o,r) x - Z

=-

(19)

where the integral is taken counterclockwise in the sense of complex integration. We leave this as an exercise and proceed to derive the analogue in R d , d 2: 3. The method below works also in R 2 . For fixed z we know that Ilx - z112-d is harmonic at all x except z. Hence so are its partial derivatives, and therefore also the following linear combination:

t j=l

2z j ~ 1 d - 2 oX j Ilx - zlld-2

1

II z l12 - IlxW Ilx-zW .

Now consider the integral: (d) zW (J fS(o,r) r2Ilx-_ IlxW

Z.

Then this is also harmonic for x E B(o, r) (why?). Inspection shows that it is a function of Ilxll alone, hence it is of the form c 1 11x11 2- d + c2 . Since its value at x = 0 is equal to r2 - d (J(r) we must have Cl = 0 and C 2 = r2 - d (J(r) = m(l). Recall the value of (J(1) from (20) of §4.2 which depends on the dimension d and will be denoted by (Ji1) in this example. Now let f be any continuous function on S(o, r) and put

f

2

1 r -llxW l(x,!) = mil) S(o,r) Ilx _ zW f(z)(J(dz).

(20)

170

4. Brownian Motion

This is the Poisson integral off for B(o, r). We have just shown that lex, 1) = 1 for x E B(o,r). We will now prove that l(x,f) = HB(x,f) for all f E qaB), so that the measure HB(x,') is given explicitly by lex, A) for all A E Iff', A caB. The argument below is general and will be so presented. Consider

-llxl12 Ilx - zlld '

r2

1

h(x,z) = mil)

then h > 0 where it is defined, and

Ilxll < r, Ilzll = r,

IS(o,r)

hex, z)a(dz) = 1. Furthermore if

Ilyll = r, y =I z, then lim x .... z hex, y) = 0 boundedly for y outside any neighbor-

hood of z. It follows easily that the probability measures lex, dz) = hex, z)a(dz) converge vaguely to the unit mass at z as x --+ z. Namely for any f E qaB), l(x,f) converges to fez) as x --+ z. Since l(',f) is harmonic in B(o, r), l(',f) is a solution to the Dirichlet problem (D,f). Hence it must coincide with HBf = HB(-,f) by Proposition 5. Let us remark that H B(X, .) is the distribution of the "exit place" of the Brownian motion path from the ball B(o, r), if it starts at x in the ball. This is a simple-sounding problem in so-called "geometric probability". An explict formula for the joint distribution of TB and X(TB) is given in Wendel

[1]. In view of the possible unsolvability of the original Dirichlet problem, we will generalize it as follows. Given D and f as before, we say that h is a solution to the generalized Dirichlet problem (D,f) iff h is harmonic in D and converges to f at every point of aD which is regular for DC. Thus we have proved above that this generalized problem always has a solution given by HDf. If D is bounded, this is the unique bounded solution. The proof will be given in Proposition 11 of §4.5. We shall turn our attention to the general question of non-uniqueness for an unbounded domain D. Consider the function 9 defined by g(x)

=

PX{T D

=

oo}

=

1 - P D c1(x).

(21)

TD < 00 almost surely so that 9 == O. In general, 9 is a solution of the generalized Dirichlet problem (D, 0). The question is whether it is a trivial solution, i.e., 9 == 0 in D. Recall from §3.6 that (for a Hunt process) a Borel set A is recurrent if and only if PAl == 1. It turns out that this identity holds if it holds in A C• For then we have for any x,

If D is bounded, then

+ EX{X(t)EAC;PX(tlTA = r{X(t) E A} + r{X(t) E A C} = 1.

r{TA < oo} ;::::PX{X(t)EA}

< oo]}

Applying this remark to A = DC, we obtain the following criterion. Proposition 7. We have 9

== 0 in D if and only if DC is recurrent.

171

4.4. Dirichlet Problem

As examples: in R d, d ~ 3, if D is the complement of a compact set, then g 1= 0 in D. In R 2 , g 1= 0 in D if and only if DC is a polar set, in which case g == 1. When g 1= 0 in D, the generalized Dirichlet problem (D,f) has the solutions H Df + cg for any constant c. The next result implies that there are no other bounded solutions. We state it in a form dictated by the proof.

Theorem 8. Let D be an open set and f be a bounded Borel measurable function on aD. If h is bounded and harmonic in D, and lim D3x --+ z h(x) = f(z) for all z E aD, then h must be of the form below: h(x) = H Df(x)

+ cg(x),

XED;

(22)

where g is defined in (21) and c is any constant. This h in fact converges to f at each point z of aD which is regular for DC and at which f is continuous. In particular if f is continuous on aD, h is a generalized solution to the Dirichlet problem (D,f). Proof. There exists a sequence of bounded regular open sets Dn such that Dn § Dn+ 1 for all n, and Dn = D. Such a sequence exists by a previous remark in the proof of Proposition 6. Suppose that h is bounded and harmonic in D. Consider the Dirichlet problem (D n , h). It is plain that h is a solution to the problem, hence we have by Proposition 5:

Un

h(x) = H Dnh(x),

(23)

This implies the first part of the lemma below, in which we write Tn for 7: Dn ' T for 7: D to lighten the typography; also To = o.

Lemma 9. For each xED, the sequence {h(X(Tn)), .?(Tn), PX; n ~ o} is a bounded martingale. We have PX-a.s.:

nli~ h(X(Tn)) = {fc(X(T)) ~~

on {T < CXl}; on {T = CXl};

(24)

where c is a constant not depending on x. Proof. Since h is bounded in D by hypothesis, we have by the strong Markov property, for each XED and n ~ 1: E X{h(X(Tn))!.?(Tn_ 1)}

=

E X (T n-ll{h(X(Tn))}

= H Dnh(X(Tn-1)) = h(X(Tn-1)) where the last equation comes from (23). This proves the first assertion, and consequently by martingale convergence theorem the limit in (24) exists.

172

4. Brownian Motion

We now prove that there exists a random variable Z belonging to the remote field '§ of the Brownian motion process, such that the limit in (24) is equal to Z almost surely on the set {T = oo}. [Readers who regard this as "intuitively obvious" should do Exercise 10 first.] Recall that '§ may be defined as follows. From each integer k ;:0: 1 let '§k be the a-field generated by X(n) for n ;:0: k, and augmented as done in §2.3, namely '§k = a(X(n), n ;:0: kr. Then'§ = /\T:= 1 '§k· By the Corollary to Theorem 8.1.4 of Course (see also the discussion on p. 258 there), '§ is trivial. [This is the gist of Exercise 4 of §4.1.] Define h in R d to be h in D, and zero in DC. We claim that there is a random variable Z (defined in Q) such that (25) for every x E Rd. For xED, under p x we have h(X(Tn)) = h(X(Tn)), so the limit in (25) exists and is the same as that ip (24). For x E DC, under p x we have h(X(Tn)) = h(x) = 0, hence the limit in (25) exists trivially and is equal to zero. Since (25) holds PX-a.s. for every x E R d , we can apply the shift to obtain for each k;:o: 1: (26) for every x E Rd. Let XED; then under p x and on the set {T = oo}, we have Tn > k and consequently h(X(Tn)) 0 f)k = h(X(Tn)) for all sufficiently large values of n. Hence in this case the limit in (26) coincides with that in (24). Therefore the latter is equal to Z 0 f)k which belongs to '§k. This being true for each k ;:0: 1, the limit in (24) is also equal to limk~oo Z 0 f)k which belongs to '§. Since '§ is trivial, the upper limit is a constant pX- a.s. for each x E Rd. This means: for each x there is a number c(x) such that PX{T

= 00; Z = c(x)} = 1.

(27)

By Exercise 4 below, g > 0 in at most one component domain Do of D, and g == 0 in D - Do. Choose any X o in Do and put for all x in D:
It is clear from the definition of Z that for any ball B = B(x, r) @ D,
4.4. Dirichlet Problem

173

have h(X(Tn)) --+ f(X(T)). This proves the first line of (24). The lemma is proved. Now it follows by (23), (25) and bounded convergence that for each

XED:

which reduces to (22) by (24). The rest of Theorem 8 is contained in Theorem 2 and is stated only for a recapitulation. Exercises 1. Without using Theorem 4, give a simple proof that a ball or a cube in R d

is regular. Generalize to any solid with a "regular" surface so that there is a normal pointing outward at each point. [Hint: (V) of §4.2 is sufficient for most occasions.] 2. Let D be the domain obtained by deleting a radius from the ball B(o, 1) in R 3 ; and let f = Ilxll on aD. The original Dirichlet problem (D,f) has no solution. 3. In classical potential theory, a bounded domain D is said to be regular iff the Dirichlet problem (D,f) is solvable for every continuous f on D. Show that this definition is equivalent to the definition given here. 4. For any open set D, show that the function 9 defined in (21) is either identically zero or strictly positive in each connected component (domain) of D. Moreover there is at most one component in which 9 > O. Give examples to illustrate the various possibilities. 5. What is the analogue of (17) and (18) for R 1 ? 6. Derive the Poisson integral in R 2 from Cauchy's formula (19), or the Taylor series of an analytic function (see Titchmarsh [1]). 7. Derive the Poisson integral corresponding to the exterior Dirichlet problem for the ball B = B(o, r); namely find the explicit distribution of X(T 8B)I{TaBO}. Compute HD(x,·), namely the distribution of X(T A ) under px, for XED. [Hint: we need the formula for Brownian motion in R 1 :r{T{oJ Edt} = Ixlj(2nt)3/2 exp[ -x 2j2t] dt; the rest is an easy computation. Answer: HD(x,dy) = r(dj2)xdn-d/21Ix - Yll-da(dy) where a is the area on A.]

174

4. Brownian Motion

9. Let D be open in Rd , d ?: 2. Then DC is polar if and only if there does not exist any nonconstant bounded harmonic function in D. [Hint: if DC is not polar, there exists a bounded continuous function f on aD which is not constant.] 10. Let {x n , n ?: I} be independent, identically distributed random variables taking the values ± 1 with probability 1/2 each; and let X n = Ij~l x j . Define for k?: 1: T k = T{k} on {Xl = +1}; T k = T{-k} on {X 1 = -I}; and A = {limk->oo X(Tk) = + CXl}. Show that each T k is optional with respect to {~n}, where ~n = O"(X j , 1 sj s n); T k i CXl almost surely; but A does not belong to the remote field I\c;;:~ 1 O"(X n , n ?: m). Now find a similar example for the Brownian motion. (This is due to Durrett.)

4.5. Superharmonic Function and Supermartingale We introduce a class of functions which stand in the same relation to harmonic functions as supermartingales to martingales. Definition. Let D be an open set. A function u is called superharmonic in D iff (a) (b)

- CXl < u s + 00; u =i= + 00 in each component domain of D; and u is lower semi-continuous in D; for any ball B(x, r) @ D we have u(x) ?:

~( ) r 0" r

JS(x.r)

u(y)O"(dy).

(1)

[A function u is subharmonic in D iff - u is superharmonic in D; but we can dispense with this term here.] Thus the sphere-averaging property in (3) of §4.3 is replaced here by an inequality. The assumption of lower semicontinuity in (a) implies that u is bounded below on each compact subset of D. In particular the integral in the right member of (1) is well defined and not equal to - 00. We shall see later that it is in fact finite even if u(x) = + 00. In consequence of (1), we have also for any B(x, r) @ D: u(x) ?: _(1)

vr

r

JB(x,r)

u(y)m(dy).

(2)

Since u is bounded below in B(x, r) we may suppose u ?: 0 there, and u(x) < 00. We then obtain (2) from (1) as we did in (6) of §4.3. It now follows that if u is superharmonic in a domain D, then it is locally integrable there. To see this we may again suppose u ?: 0 because u is bounded below on each compact

175

4.5. Superharrnonic Function and Superrnartingale

subset of D. Let A be the set of points x in D such that u has a finite integral over some ball with center at x. It is trivial that A is open. Observe that if x E A, then there is X n arbitrarily near x for which u(x n ) < 00, and consequently U has a finite integral over any B(x m r) § D. Therefore this is also true for any B(x, r) § D by an argument of inclusion. Now let X n E A, X n ~ xED. Then for sufficiently large n and small r we have x E B(x n , r) § D. Thus x E A and so A is closed in D. Since D is a domain and A is not empty we have proved A = D as claimed. We now follow an unpublished method of Doob's to analyse superharmonic functions by means of supermartingales. Suppose first that D is bounded and u is superharmonic in an open set containing 15. Let p denote the distance function in Rd. For a fixed 8 > 0 and each x in D we put r(x)

= !p(x, aD) /\ 8,

S(x)

=

S(x,r(x));

thus S(x) is the sphere with center x and radius r(x) varying with x as prescribed. Define a sequence of space-dependent hitting times {Tn, n ::?: O} as follows: To = 0 and for n ::?: 1: Tn

= inf{t >

Tn-1IX(t)

E

S(X(Tn- 1))}

(3)

where inf 0 = + 00 as usual. Each Tn is optional with respect to {~} (exercise !). The strong Markov property of X and the inequality (1) imply the key relation:

(4) provided that u(x) < 00. The integrability of u(X(Tn )) is proved by induction on n under this proviso. Thus {u(X(Tn)), $'(Tn), n::?: I} is a supermartingale under P Since D is bounded all the terms of the supermartingale are bounded below by a fixed constant. Hence it may be treated like a positive supermartingale; in particular, it may be extended to the index set "1 :::;; n :::;; 00" in a trivial way. Let us first prove the crucial fact that for each xED, we have X

•

X

p {lim Tn = n~oo

LD} = 1.

(5)

For it is clear that PX-a.s., Tn 1 and Tn < L D < 00 because D is bounded. Let Too = limn 1 Tn" Then X(T (0) = limn~oo X(T n) by continuity and so X(T (0) E 15. Suppose if possible that X(T (0) E D. Then for all sufficiently large values of n, we have p(X(Tn- 1), aD) > r1p(X(T (0)' aD) > Oandp(X(Tn_ 1), X(Tn))< 4- 1p(X(Too ),aD)/\8. But by (3), p(X(Tn-1),X(Tn)) = 2- 1p(X(Tn _ 1),aD)/\8. These inequalities are incompatible. Hence X(T (0) E aD and consequently Too =

LD'

176

4. Brownian Motion

Now fix t> 0 and define a stopping random variable as follows: N = inf {n ;;:::

11 Tn ;;::: t}.

(6)

where inf 0 = + 00. For each integer k ;;::: 1, {N = k} belongs to the (J-field generated by T 1 , . . . , T k ; hence to $'(Tk ). Thus N is optional with respect to {$'(Tn), n;;::: I}. Therefore by Doob's stopping theorem for a positive (hence closable) supermartingale (Theorem 9.4.5 of Course):

(7) is a supermartingale under Px provided u(x) < = $'(T v) by (5). It follows that

00.

Here X(T exJ = X(TD),

$'(T ex,)

by Fatou's lemma since all the terms are bounded below. Now limn X(TnAN ) = X(TN) whether N is finite or infinite, and by the lower semicontinuity of u in l5 (because u is superharmonic in an open set containing l5) we conclude that (9)

where we have indicated the dependence of N on s. On the set {t < Tv}, there exists n such that t < Tn; hence N(s) < 00 and T N(e) _ 1 < t :::;; T N(e) by (6). By the definition of T N(e), this implies IX(t) - X(TN(e)) I:: ; S. Thus we have proved lim X(TN(e)) = X(t 1\ T D)

(10)

ei 0

PX-a.s.on {t < Tv}. On theset{t;;::: Tv},N(s) = + 00 for all s > oand X(TN(e)) = X(T v ) by (5). Hence (10) is also true. Using (10) in (9) and the lower semicontinuity of u again, we obtain (11)

The next step is standard: if 0 < s < t, we have PX-a.s.

E u(X(t 1\ Tv)) I~} = EX(s){ u(X((t - s) 1\ TD))} :::;; u(X(s)) = u(X(s 1\ TD)) X

{

(12)

on {s < Tv}, since X(s) E D; whereas on {s ;;::: TD} the two extreme members of (12) both reduce to U(X(TD)) and there is equality. We record this important result below.

177

4.5. Superharmonic Function and Superrnartingale

Theorem 1. If D is bounded and u is superharmonic in an open set containing

15, then {u(X(t 1\ TD)), ff(t), 0:<:;; t:<:;; oo}

(13)

is a supermartingale under p x for each XED for which u(x) < 00. When u is harmonic in an open set containing 15, then (13) is a martingale under p x for each XED. We proceed to extend this result to an unbounded D and a function u which is superharmonic in D only. At the same time we will prove that the associated supermartingale has right continuous paths with left limits. To do this we use the basic machinery of a Hunt process given in §3.5. First, we must show that the relevant process is a Hunt process. For the Brownian motion in R d this is trivial; indeed we verified that it is a Feller process. But what we need here is a Brownian motion somehow living on the open set D, as follows (see Exercise 9 for another way). Let D be any open subset of R d , and Do = D u {a} the one-point compactification of D. Thus a plays the role of "the point at infinity": x --+ a iff x leaves all compact subsets of D, namely x --+ aD; and all points of aD are identified with a. Define a process {X(t); t ~ O} living on the state space Do as follows:

X(t) = {X(t),

a,

~f t < T D;

IfTD:<:;;t:<:;;oo.

(14)

By this definition ais made an absorbing state, and T D the lifetime of X; see (13) and (14) of §1.2. Define a transition semigroup (Qt, t ~ 0) as follows; for any bounded Borel function