Some applications of the Radon-Nikodymtheorem to asymptotic martingales

Abtin Daghighi

U.U.D.M. Project Report 2006:1

Examensarbete i matematisk statistik, 20 poäng

Handledare och examinator: Allan Gut

Mars 2006

Department of Mathematics

Uppsala University

Some applications ofthe Radon–Nikodym theorem

to asymptotic martingalesSupervisor Professor Allan Gut

Author Abtin Daghighi

Abstract

The Radon–Nikodym theorem for signed measures is presented, and itsconnection with asymptotic martingales is investigated. Weak convergenceof vector valued (E-valued) asymptotic martingales such that E possessesthe Radon–Nikodym property and has a separable dual, is proved. Atheorem stating that strong, almost everywhere convergence of Banachspace valued asymptotic martingales is equivalent to the requirement thatthe Banach space has finite dimension is proved.

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AcknowledgmentTremendous thanks to the supervisor, Professor Allan Gut, for professional andhumble help and guidance. Enormous thanks to Professor Svante Janson whohas spent precious time to clear the paper from its many flaws. Many thanks toProfessor Christer Kiselman who has tried to teach the author to work throughand review mathematical texts. Also thanks to the head of computer networkcoordination at the department of mathematics in Uppsala, Mr Carl Edstrom,and head of briefing, Mrs Inga-Lena Assarsson, for administrative guidance.

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Contents1 Introduction 4

2 Preliminaries 4

3 Statement of the Radon–Nikodym theorem 6

4 Some results required for the proofs of the Radon–Nikodym the-orem 6

5 Two different proofs of the Radon–Nikodym theorem 135.1 A first proof of the Radon–Nikodym theorem . . . . . . . . . . . . 135.2 A second proof of the Radon–Nikodym theorem . . . . . . . . . . 14

6 Conditional probability 15

7 Martingales 17

8 Asymptotic martingales 198.1 Definition and brief theory . . . . . . . . . . . . . . . . . . . . . . 198.2 The Radon–Nikodym property for vector-valued asymptotic mar-

tingales . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 248.3 Statement of the main theorem . . . . . . . . . . . . . . . . . . . 258.4 Remark on a strong convergence result for martingales . . . . . . 268.5 Proof of the main theorem . . . . . . . . . . . . . . . . . . . . . . 27

9 References 31

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1 IntroductionIn this paper we review some important fields in probability theory, namely con-ditional probability, martingales and asymptotic martingales. We shall then in-vestigate the use of the Radon–Nikodym theorem in the theory of asymptoticmartingale convergence.

The Radon–Nikodym theorem provides conditions for existence of a certain in-tegral representation of measures. The paper first presents and proves the Radon–Nikodym theorem. This theorem was first proved for Rn by Johann Radon (1847-1956) in 1913 and was generalized by Otto Marcin Nikodym (1887-1974) in 1930(cf. references). Throughout this paper the theorems that are required for theproofs of the main results shall be presented and proved separately. The litera-ture used for the presentation of the Radon–Nikodym theorem is Rana (1997),Halmos (1950) and Cohn (1996). Some general facts from probability theoryshall be presented followed by an application concerning the definition of condi-tional probability. Then some basic martingale theory will be presented. Thetheory of asymptotic martingales shall be reviewed briefly and an application ofthe Radon–Nikodym theorem to this theory shall be presented in the form of atheorem concerning weak convergence of vector-valued asymptotic martingales,on a Banach space (complete normed space) which posesses the Radon–Nikodymproperty and has a seperable dual. We shall then conclude with the main theoremof the paper which states that strong almost everywhere convergence of Banachspace (E) valued asymptotic martingales is equivalent to E having finite dimen-sion. The most important reference literature which has been used to reviewthe main asymptotic martingale applications is due to Alexandra Bellow. Forthe understanding of the theory of asymptotic martingales, which is required inorder to be able to appreciate the of the content of these convergence theorems,the important educative work of Gut (1983) has been used.

2 PreliminariesThe reader is assumed to have taken an introductory course in measure theory.This means that we shall in the text refer to some well–known theorems withoutproof. These include: The monotone convergence theorem, Fatou’s lemma andthe Lebesgue bounded convergence theorem. For the proofs the reader is referredto a textbook in measure theory e.g. Halmos (1950) or Cohn (1996).

Some results required in order to prove the theorems shall be stated and provedseparately, before the proof of the main theorem. In order to show a differentand perhaps more modern approach in proving the Radon–Nikodym theorem,the paper begins by presenting a proof which uses the Lebesgue decompositiontheorem and the Riesz representation theorem.

We shall begin by giving a few definitions in order to accustom the reader tothe letters and symbols that are associated to the concepts in this paper.Definition.

A measure space, denoted (X;S; ), is a measurable space together with a

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measure, , on S. A set E is called positive (negative) if 8F such that F ismeasurable E \ F is also measurable and (E \ F ) 0 ((E \ F ) 0).Definition. A signed measure is an extended real valued, countably additiveset function on the class of all measurable sets of a measurable space (X;S);such that (;) = 0 and assumes at most one of the values +1 and 1 (inparticular, if fEngn is a disjoint sequence then

P1n=1 (En) is either convergent

or definitely divergent to +1 or 1).

Definition. Let (X;S) be a measurable space and ; be signed measures onS. The measure is called absolutely continuous with respect to , denoted as , if (E) = 0 for every measurable set E for which jj(E) = 0:

Definition. A simple function f =Pn

i=1 aiIEi(x); where the Ei are disjoint, iscalled integrable if (Ei) <1 for all i for which ai 6= 0.

Definition. A nonnegative function f is called S-measurable if there exists anincreasing sequence, fsngn, of nonnegative simple functions such that f(x) =limn sn; for all x 2 X:

Definition. A function f on a measure space (X;S; ) is integrable if there existsa mean fundamental sequence ffng of integrable simple functions which convergein measure to f . The integral of f , denoted

Rfd, is defined byZ

fd = limn!1

Zfnd:

We now give the statements of the three important theorems which the readeris required to know from a basic course in measure theory:The monotone convergence theorem. Let ffng be an increasing sequence ofextended real valued nonnegative measurable functions such that limn!1 fn(x) =f(x) almost everywhere. Then f is a nonnegative measurable function andlimn!1

Rfnd =

Rfd:

Fatou’s lemma. If ffng is a sequence of nonnegative measurable functions,then Z

lim infn!1

fnd lim infn!1

Zfnd:

The Lebesgue dominated convergence theorem. If ffng is a sequence ofintegrable functions which converges in measure () to f , and if g > 0 is anintegrable function such that jfn(x)j g(x) almost everywere, n = 1; 2; : : : ; thenf is integrable and limn

Rfnd =

Rfd:

In probability theory the set we have often referred to above as X is denotedby Ω and F denotes a -algebra (of subsets on Ω):Definition. A probability measure, denoted P ; on F is a real valued functionwith domain F such that:

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(i) 8e 2 F ;P(e) 0;

(ii) P(Ω) = 1:

(iii) If feng is a countable family of disjoint sets in F then P (S

n en) =P

nP(en);

We denote by B (the Borel -algebra on R) the minimal -algebra containingthe collection of intervals f(a; b] : 1 < a < b <1g:

The measurable space (Ω;F ;P) is called a probability space. A real extendedvalued random variable is a function X with range in [1;+1] such that foreach b 2 R [ f+1 or 1g,

f! 2 Ω: X(!) bg 2 F :

3 Statement of the Radon–Nikodym theoremThe theorem known as the Radon–Nikodym theorem for signed measures gives arepresentation involving a measurable function for signed measures.The Radon–Nikodym Theorem. Let (X;S; ) be a -finite measure space. Ifa -finite signed measure on S is absolutely continuous with respect to , thenthere exists a finite real–valued measurable function f on X such that either f+

or f is integrable and

(E) : =ZE

fd =ZE

f+d +ZE

fd

for every measurable set E. Furthermore, if (E) =REgd, E 2 S; then fx 2

X : f(x) 6= g(x)g is a set of measure zero with respect to jj, i.e., f is uniquealmost everywhere with respect to .

The function f is sometimes denoted dd

, and called the Radon–Nikodymderivative.

4 Some results required for the proofs of theRadon–Nikodym theorem

In this chapter we present some of the theorems and propositions whose resultswill be used in the proofs of the Radon–Nikodym theorem. We refer to Rana(1997), Halmos (1950) and Cohn (1996).

We begin with a decomposition theorem that for any measure space (X;S; )with a signed measure ; provides two disjoint sets such that is positive respec-tively negative on these sets.The Hahn decomposition theorem. If is a signed measure, then for themeasure space (X;S; ) there exist two disjoint sets A and B such that A ispositive and B is negative with respect to and such that X = A [B.

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Proof. The measure is a signed measure so w.l.o.g. assume that for everymeasurable set E, 1 < (E) 1: Next it is noted that any countable unionof negative sets is a negative set, since the difference of two negative sets isnegative and since a countable disjoint union of negative sets is again a negativeset. Denote the infimum of all measurable negative sets b as = inf (b), and letfbig be a sequence of measurable negative sets such that limi!1 (bi) = . Bythe argument above B =

S1i=1 bi is a measurable negative set and furthermore

such that (B) is minimal.Now consider the set X rB and assume that this set is not positive in order

to arrive at a contradiction.Let E0 be a measurable subset of X r B such that (E0) < 0 (such a set

exists if X r B is not positive). Then E0 cannot be a negative set since then(B [E0) < (B), which would contradict the fact that (B) was minimal. Letk1 be the smallest positive integer such that there is a measurable set E1 E0with the property (E1) 1

k1((E1) will be finite since (E0) < 0). Now since

(E0 r E1) = (E0) (E1) (E0) 1k1

< 0;

the argument which has been applied to E0 may be applied to E0rE1 also. Thuschoosing k2 as the smallest positive integer such that there is a measurable setE2 E0rE1 with the property (E2) 1

k2, and so on. This describes a procedure

that can be iterated to infinity. Furthermore Ej+1 E0 nSj

i=1 Ei, thus the Ei

are disjoint.Now for any measurable subset G E0, j(G)j < 1: This is true since

j(E0)j <1, and (E0) = (E0 rG) + (G) which means that if exactly one orboth of (E0 r G) and (G) are infinite then also (E0) must be infinite (since assumes at most one of the values +1 and 1). Thus the only remainingpossibility is that (E0 rG) and (G) are both finite.

This means that limn!1 1kn

= 0, since we can set G =S1

j=1 Ej, which is acountable union of measurable subsets of E0, and 1 > (G) =

Pn

1kn: Define

F0 = E0 r

1[j=1

Ej:

Then, by construction, F0 must be a negative set since if there existed a positivesubset E such that (F0rE) (E) then Ei would already have been includedinS1

j=1 Ej.

(F0) = (E0)1Xj=1

(Ej) (E0) < 0;

so that (F0 [ B) < (B); which again contradicts the fact that B is minimal.Hence X rB is a positive set and by choosing A = X rB the conditions of thetheorem are satisfied. This completes the proof.

In connection with the Hahn decomposition the concept of total variation ofa measure may be defined.

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Definition. Given X;A;B; as in the theorem, define for any measurable set Ethe following set functions

+(E) = (E \ A); (E) = (E \B):

+ and are called the upper and lower variation of . The total variation of, denoted jj; is then defined according to

jj(E) = +(E) + (E):

Next we shall prove a theorem called the von Neumann theorem which willbe a prerequisite in proving the so called Lebesgue decomposition theorem whichin turn directly implies the existence part of the proof of the Radon–Nikodymtheorem for -finite measures. In the proof of the von Neumann theorem we willneed Proposition 1 below and the Riesz representation theorem.

The Riesz representation theorem. Let T : L2(X;B; ) ! C (where Cdenotes the set of complex numbers) be a bounded linear functional (where weuse the notation Lp(X;B; ) for the set of S-measurable functions, f , such thatR jf jpd <1). Then there exists a unique g0 2 L2(X;S; ) such that

T (f) = hf; g0i for all f 2 L2(X;S; ):

Proof. Denote the nullspace of T by Ker(T ): If there exists a g0 2 L2(X;S; )such that T (f) = hf; g0i, then g0 2 (Ker(T ))?; where Ker(T ) := ff 2 L2(X;S; ): T (f) = 0g L2(X;S; ): This is true since T is linear. Also, T is continuousso Ker(T ) is closed. Now, if Ker(T ) = L2(X;S; ); then T (f) = 0 for all f 2L2(X;S; ) and T (f) = hf; 0i 8f 2 L2(X;S; ). If instead T (f) 6= 0 for somef 2 L2(X;S; ); then (Ker(T ))? 6= 0: Let 0 6= g 2 (Ker(T ))?: We have that

T (g) 6= 0

since if g 2 KerT , then g ? g. We will find the desired g0 as follows: Define g0 :=T (g)hg;gi

g: For all z 2 C; T (zg) = T (g)z and hzg; g0i = zhg; g0i = zhg; T (g)

hg;gigi =T (g)z: Hence

(4.1) T (zg) = hzg; g0i; 8z 2 C:

Also, if f 2 Ker(T ); then

(4.2) T (f) = 0 = hf; g0i;and if f =2 Ker(T ); then (f zg) 2 Ker(T ) iff

0 = T (f zg) = T (f) zT (g);

which gives z = T (f)T (g) : Hence for f =2 Ker(T ); it follows that

f g T (f)

T (g)

2 Ker(T )

and therefore, by (4:1) and (4:2); that

T (f) = T

f g

T (f)T (g)

+ T

gT (f)T (g)

= hf g

T (f)T (g)

; g0i+ hgT (f)T (g)

; g0i = hf; g0i

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(where (4:1) has been applied to the term Tg T (f)T (g)

and (4:2) has been applied

to the term Tf g T (f)

T (g)

). This proves existence.

Now, if hf; g0i = hf; g1i 8f 2 L2(X;S); then hf; g0g1i = 0 8f 2 L2(X;S):In particular, letting f = g0g1 gives 0 = kg0g1k, so that g0 = g1; which provesuniqueness. This completes the proof.

Proposition 1. Let f 2 L1(X;S; ); < 1; Y C; Y be closed. If for anyE 2 S such that (E) > 0 we have

1

(E)

REfd

2 Y; then f(x) 2 Y for almost

every x 2 X with respect to .

Proof. It suffices to show that (fx : f(x) =2 Y g) = 0: Since Cr Y = Y c is open,there exist countably many open balls fBngn1 such that Y c =

S1n=1 Bn: Thus,

fx 2 X : f(x) =2 Y g = fx 2 X : f(x) 21[n=1

Bng =1[n=1

f1(Bn):

But f is measurable and we have f1(E) 2 S for every Borel set E X; and theopen balls, Bn, are Borel sets. Now fix an integer n and let Bn have center z0and radius r0: Let En = f1(Bn) and suppose that (En) > 0; in order to arriveat a contradiction. Then

j 1(En)

ZEn

fd z0j = j 1(En)

ZEn

(f(x) z0)dj

1(En)

ZEn

jf(x) z0jd r0;

which implies that 1(En)

REn

fd 2 Bn Y c; which is a contradiction to theassumption 1

(E)

REfd 2 Y: Hence (En) = 0; which proves the statement.

The von Neumann theorem. Let ; be two -finite measures on a measur-able space (X;S). Then there exist mutually disjoint sets Xi 2 S; 1 i 3; suchthat the following hold:

(i) X =S3

i=1 Xi;

(ii) (X3) = (X1) = 0;

(iii) There exists a nonnegative measurable function g on X such that g(x) >08x 2 X2; and such that for all E 2 S with E X2 we have

(E) =ZE

gd:

Proof. First consider the case where both and are finite. Then + (which isdefined by (+)(E) = (E) +(E); E 2 S) is a finite measure and L2(+) L1( + ) L1(): This is true since the function h 1 belongs to L2( + )

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and by the Cauchy–Schwarz inequality,Zjf jd( + ) (( + )(X))1=2

Zjf jd( + )

1=2

:

Now, if f 2 L1(+ ) thenR jf jd R jf jd(+ ) which gives f 2 L1(): Define

T (f) : L2( + ) ! R as

T (f) =Z

fd; f 2 L2( + ):

Then T is a bounded linear functional since jT (f)j ((X0))1=2kfk2: By theRiesz representation theorem there exists a function f0 2 L2( + ); such thatT (f) =

Rff0d( + ); 8f 2 L2( + ); so thatZ

fd =Z

ff0d( + ):

Putting f = IE; E 2 S where IE is the indicator function for E, gives

(4.3) (E) =ZE

f0d( + );

for every E 2 S:

Using Proposition 1 with Y = R0; f = f0; and replaced by ( + ) gives

(4.4) ( + )fx 2 X : f0(x) =2 R0g = 0:

Thus f0 is real-valued (so f0= f0) for almost every x 2 X and f0(x) 0 foralmost every x with respect to (+ ): Also (+ )(E) (E) =

REf0d(+ );

for all E 2 S; which implies thatRE

(1 f0)d( + ) 0: Using Proposition 1again but with the closed interval Y = [0; 1]; gives 0 f0 1 for almost everyx 2 X with respect to ( + ): LetN := fx 2 X : f0(x) > 1g [ fx 2 X : f0(x) < 0g;X 0

1 := fx 2 X : f0(x) = 1g; X1 = X 01 [N;

X2 := fx 2 X : 0 < f0(x) < 1g;X3 := fx 2 X : f0(x) = 0g:Then ( + )(N) = 0 and Xi; i = 1; 2; 3 are pairwise disjoint with X =

S3i=1 Xi:

Furthermore, (X3) =RX3f0d( + ) = 0 and

(X 01) = ( + )(X 0

1) (X 01) =

ZX0

1

(1 f0)d( + ) = 0:

(N) = 0 then implies (X1) = 0. Also (E) =REf0d(+ ); for all E X2; so

that (E) REf0d =

REf0d: ThusZ

E

(1 f0)d =ZE

f0d; 8E 2 X2:

Therefore, ZX2

s(1 f0)d =ZX2

sf0d;

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for every nonnegative simple measurable function s: Now we know that 0 f0 1and since for every nonnegative measurable function f its integral is defined asthe limit of integrals of nonnegative simple measurable functions fsigi we get

limi!1

ZX2

si(1 f0)d =ZX2

f(1 f0)d =ZX2

ff0d:

For E X2; define

f(x) :=

(IE(x)

1f0(x) ; x =2 X1;

0; otherwise.

Then 8E 2 S with E X2; (E) =RE

f0(x)1f0(x)d: Now define

g(x) :=

(f0(x)

1f0(x) ; x 2 X2;

0; otherwise.

Then g is a nonnegative measurable function and, 8E 2 S with E X2; we have

(E) =ZE

gd:

This proves the theorem for the case of finite measures.For the case where ; are -finite measures one can write X =

S1i=1 Yi; where

the sets Yi 2 S are pairwise disjoint, (Yi) < 1 and with (Yi) < 18i 1: Bythe proof of the finite case above one can find, for every i 1; mutually disjointsets X1

i ; X2i ; X

3i 2 S; such that Yi = X1

i [ X2i [ X3

i with (X1i ) = (X3

i ) = 0:Also, for every i 0; there exists a measurable function gi such that gi(x) > 0for almost every x 2 X2

i and gi = 0 for almost every x 2 X1i [X3

i : Furthermore(E \X2

i ) =R

(E\X2i ) gid; for all E 2 S: Now define

X1 :=S1

i=1 X3i ; X2 :=

S1i=1 X

2i ; X3 :=

S1i=1 X

3i ; and

g : =

(gi(x); x 2 X2 for some i;0; otherwise.

Then X1; X2; X3 and g satisfy the requirements of the theorem. This completesthe proof.

The Lebesgue decomposition theorem. Let ; be two -finite measureson a measurable space (X;S). Then there exist -finite measures a and s withthe following properties:

(i) = a + s,

(ii) There exists a nonnegative measurable function f such that

a(E) =ZE

fd;

for every E 2 S:

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(iii) There exists a set A 2 S such that (Ac) = s(A) = 0:

Furthermore such a decomposition is unique.Proof. By the von Neumann theorem there exist disjoint sets X1; X2; X3 in Ssuch that X = X1 [ X2 [ X3; and (X3) = (X1) = 0: Also, there exists anonnegative measurable function g on X with the properties g(x) > 0 on X2 andg(x) = 0 on Xc

2, such that

(E \X2) =ZE\X2

gd; 8E 2 S:

Denote A := X2 [X3; and define, for all E 2 S; a(E) := (A\E) and s(E) :=(E \ X1): Then = a + s; and s(A) = (Ac) = 0: Moreover a(E) =(E \ (X2 [X3)) = (E \X2) =

RE\X2

gd: Define

f :=

(g(x); x 2 X2;

0; x 2 X1 [X3:

Then f is a nonnegative measurable function on X and

a(E) =ZE

fd; 8E 2 S:

This proves statements (i) and (ii): It remains to prove uniqueness. Assume thereexist measures 0a and that 0s; a set A0 and a nonnegative measurable function f 0

such that = 0a + 0s; ((A0)c) = 0s(A0) = 0; 0a(E) =REf 0d;8E 2 S:

Then((A0 \ A)c) = (A0c [ Ac) = 0 = s(A0 \ A) = 0s(A0 \ A):

Furthermore,

(4.5) a((A0 \ A)c) = 0a((A0 \ A)c) = 0;

since the measure of the given set with respect to is zero. Since a(E)+s(E) = 0a(E) + 0s(E); it follows that

(4.6) a(E) 0a(E) = s(E) 0s(E);

8E 2 S such that (E) < +1: Since f is nonnegative and E \ (A0 \ A)c is asubset of a set of measure zero we get

(4.7) a(E \ (A0 \ A)c) 0a(E \ (A0 \ A)c) = 0:

We now note that a(E) = a(E \ (A0 \ A)c) + a(E \ (A0 \ A)); and 0a(E) =a(E \ (A0 \A)c) + 0a(E \ (A0 \A)): Making this substitution in (4:6) and thenapplying (4:5) gives

a(E \ (A0 \ A)) 0a(E \ (A0 \ A)) = s(E \ (A0 \ A)) 0s(E \ (A0 \ A)) = 0:

Thusa(E \ (A0 \ A)) = 0a(E \ (A0 \ A));

8E 2 S with (E) < +1; which together with (4:5) shows that a(E) =

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0a(E); 8E 2 S such that (E) < +1: Since is -finite this means thata(E) = 0a(E): An analogous argument for s gives s(E) = 0s(E): This com-pletes the proof.

The second proof of the Radon–Nikodym theorem given in this paper is ratherconventional. In the proof we shall need the following proposition:Proposition 2. If and are finite measures such that and is notidentically zero, then there exist > 0, and a measurable set A such that (A) > 0and such that A is a positive set for the signed measure .Proof. Let X = An [ Bn be a Hahn decomposition with respect to the signedmeasure

n; n = 1; 2; : : :, and let

A0 =1[n=1

An; B0 =1\n=1

Bn:

B0 Bn then gives 0 (B0) 1n(B0), n = 1; 2; : : :, which implies that

(B0) = 0. Since by assumption is nonzero we have (A0) > 0 and therefore,by absolute continuity, (A0) > 0. This implies that (An) > 0 for at leastone value of n, say N . Hence one may choose A = AN and = 1

Nso that the

requirements of the theorem are satisfied. This completes the proof.

5 Two different proofs of the Radon–Nikodymtheorem

This chapter will present two different proofs of the Radon–Nikodym theorem.The proofs are given in separate sections and will use different theorems andpropositions from the preceding chapter.

5.1 A first proof of the Radon–Nikodym theoremIn this section we shall use important theorems from the preceding chapter togive a proof of the Radon–Nikodym theorem (we refer to Rana (1997)):Proof of the Radon–Nikodym Theorem for signed measures. We firstbegin with the case where and are -finite (hence not signed).

Since , a = and s = 0 in the Lebesgue decomposition theoremand by the same theorem there is a nonnegative measurable function f such thata(E) =

REfd; for every E 2 S: To prove uniqueness assume existence of g 6= f

as in the statement. Suppose there exists a set E 2 S such that (E) > 0 andf(x) > g(x) 8x 2 E: Since ; are -finite, we can choose A 2 S such that(A) <1; (A) +1 and (E \ A) > 0: ThenZ

A\E(f(x) g(x))d(x) = (A \ E) (A \ E) = 0;

which is a contradiction. Thus f(x) g(x) for almost every x 2 X. This provesthe theorem for the case of -finite measures.

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Now let be signed. We shall next use the following statement:

() iff + and ;

Proof of the statement. Let A;B be a Hahn decomposition of X with respect to (and + , the associated positive and negative parts). Let (E) = 0. Thus(A \ E) = (B \ E) = 0, and by hypotesis (A \ E) = (B \ E) = 0. Thus+(E) = (A \ E) = 0 and (E) = (B \ E) = 0. This proves ().Since is -finite the measures + and are finite (non-signed) measures.Then applying our proof for this case to each of these measures completes theproof of the theorem for signed measures since = + + .

5.2 A second proof of the Radon–Nikodym theoremThis section shall give a more conventional proof of the main theorem. We referto Halmos (1950). The proof begins by considering the case of finite measuresand for this case it turns out that the function f actually becomes integrable,which is important for the applications of the theorem that are discussed in theforthcoming chapters.A second proof of the Radon–Nikodym theorem for signed measures.Consider the case where both and are finite measures. Let K be the classof all nonnegative functions, f , that are integrable with respect to , such thatREfd (E) for every measurable set E. Furthermore let

= sup

(Zfd : f 2 K

):

Let ffng be a sequence of functions in K such that

limn!1

Zfnd = :

Then set gn = f1 _ _ fn (or maxffi; 1 i ng) which will be nonnegativesince all fn are nonnegative. Then any measurable set E may be written as afinite, disjoint union of measurable sets, E = E1[ [En, so that gn(x) = fj(x),x 2 Ej; for some j = 1; : : : ; n. This givesZ

E

gnd nXj=1

Zfjd

nXj=1

(Ej) = (E) <1;

since we have assumed ; to be finite. This shows that fgngn is an increasing se-quence of measurable functions in K. Now let f0(x) = supffn(x) : n = 1; 2; : : :g:Then f0(x) = limn!1 gn(x). By the monotone convergence theorem f0 is mea-surable and since we have assumed ; to be finite f0 is also integrable. FurtherRE

limn!1 gnd (E) for every measurable set E gives f0 2 K andRf0d = :

Since f0 is integrable, there exists a finite valued function, which we have denotedby f in the statement of the theorem, such that f0 = f almost everywhere withrespect to .

14

Now consider the measure defined by

0(E) = (E)ZE

fd:

This measure must be identically zero. To prove this assume that it is not iden-tically zero to get a contradiction. By Proposition 2 there exists > 0 and ameasurable set A such that (A) > 0 and such that

(E \ A) 0(E \ A) = (E \ A)ZE\A

fd;

for every measurable set E. If g = f + IA, thenZE

gd =ZE

fd + (E \ A) ZErAc

fd + (E \ A) (E);

for every measurable set E. Thus g 2 K. ButZE

gd =ZE

fd + (A) > ;

which is a contradiction to the maximality of f . Hence the assumption that 0 isnot identically zero does not hold.

This thus proves the existence of f such that (E) =REfd, for the case of

finite and . Now f is integrable since is finite. Assume that there exists f1which also fulfills the requirements of the theorem. Then

RF

(f f1)d = 0; forevery measurable set F: Let E = fx : f(x) f1(x) > 0g, then

RE

(f f1)d = 0.This implies that (E) = 0. To see that this is true, let En = fx : (f(x)f1(x)) 1ng, n = 1; 2; : : :. Now

0 =ZE\En

(f f1)d 1n(E \ En) 0:

But since E =S1

n=1 En, the relation (E) P1n=1 (En \ E) implies (E) = 0.

Replacing (f f1) with (f1 f) shows that fx : (f(x) f1(x)) < 0g is also a setof measure zero. This proves uniqueness.

Finally, X is a countable, disjoint union of measurable sets Ek; k 2 A, where Ais a countable index set, on which ; are finite. Therefore writing f =

Pk2A fkIk

where Ik denotes the indicator function for Ek one gets thatRf =

Pk

Rfkd:

Using the statement () from the first proof we have

() iff + and :

Now we can write (E) :=REfd =

REf+d R

Efd = + + and since

is signed at least one of the integrals must be finite. This completes the proof.

6 Conditional probabilityIn this chapter an application of the Radon–Nikodym theorem to probabilitytheory shall be presented. The result is found in, e.g., Gut (2005) or Chung

15

(1968) and concerns the definition of conditional probability. We begin with adefinition.

Definition. If Λ is a set in F such that P(Λ) > 0; then define the conditionalprobability relative to Λ as follows

PΛ(E) = P(Λ \ E)P(Λ)

:

Further, the conditional expectation relative to Λ is

EΛ(Y ) =Z

ΩY (!)PΛ(!) = 1

P(Λ)

ZΛY (!)P(!):

Let G be the Borel -field generated by a countable partition fΛng whereΛn = fX = ang; X a discrete random variable. Given an integrable randomvariable Y define a discrete random variable by

E(Y jG) =Xn

EΛn(Y )IΛn

;

where IΛnis the indicator function. This variable takes the value EΛn

(Y ) on theset Λn: Thus

E(Y ) =1Xn=1

ZΛn

Y (!)PΩ(!) =1Xn=1

P(Λn)EΛn(Y )

=1Xn=1

ZΛn

E(Y jG)dP =Z

ΩE(Y jG)dP :

Further, any Λ 2 G is the union of a subcollection of the Λn, sayS

j2JN Λj;

which givesR

Λ Y dP =P

j

RΛjY dP =

Pj

RΛjE(Y jG)dP =

RΛ E(Y jG)dP : Thus

(6.1) 8Λ 2 G :Z

ΛY dP =

ZΛE(Y jG)dP :

Now assume that there exist two functions '1; '2 such that

8Λ 2 G :Z

ΛY dP =

ZΛE(Y jG)dP =

ZΛE'idP ; i = 1; 2:

Then '1 = '2 almost everywhere since f! : '1(!) > '2(!)g 2 G and over thisset the integral of 1 '2 cannot be zero unless the set has probability zero (i.e.zero measure). Similarly for the set f! : '1(!) < '2(!)g 2 G. This provides uswith equivalence classes of random variables, namely the integrands in (6:1). Wedenote the corresponding equivalence classes by E(Y jG), and we call a memberof such a class a version of the conditional expectation.

Now we are ready to prove the following uniqueness theorem.

Theorem (conditional expectation). If E(jY j) < 1 and G is a sub--fieldof a -field F , then there exists a unique equivalent class of random variables

16

E(Y jG) belonging to G such that the following holds

(6.2) 8Λ 2 G :Z

ΛY dP =

ZΛE(Y jG)dP :

Proof. Consider the set function on G:

8Λ 2 G : (Λ) =Z

ΛY dP :

It is finite valued and countably additive, hence a signed measure on G: If P(Λ) =0 then (Λ) = 0; hence P . Then by the Radon–Nikodym theorem thetheorem is proved and we have that E(Y jG) coincides with the derivative d

dP :

The theorem shows existence and uniqueness of conditional expectation whichcan hence be defined as follows:Definition. Given an integrable random variable Y and a Borel subfield G; theconditional expectation E(Y jG) of Y relative to G is any member of the equivalenceclass of random variables on Ω such that

(i) it belongs to G,

(ii) it has the same integral as Y over any set in G.

7 MartingalesIn this chapter a brief introduction shall be given to the theory of martingales.The purpose is to facilitate the presentation of asymptotic martingales and theirconnection to the Radon–Nikodym theorem. The literature used for this reviewis Gut (2005), Chapter 10, and Ross (1996). We begin with a few definitionsconcerning sequences of random variables and associated fields.

Consider the probability space (Ω;F ;P): Let fFn; n 0g be a sequence ofsub--algebras of F such that F0 F1 Fn Fn+1 F . Then thesequence is called a filtration. The following lemma will be used in calculatingthe expectation of a certain type of sequences of random variables later in thischapter.The smoothing lemma. If F1 F2; then

E(E(XjF2)jF1) = E(XjF1) = E(E(XjF1)jF2);

almost surely.Proof. Let Λ 2 F1 F2: ThenZ

ΛE(E(XjF2)jF1)dP =

ZΛE(XjF2)dP =

ZΛXdP =

ZΛE(XjF1)dP ;

where (6:1) has been used for the first two equalities. This proves the first equalityof the theorem. Now F1 F2 implies that E(XjF1) 2 F2: But then the secondequality of the theorem holds by (6:1): This completes the proof.

17

Next we give a definition that will be used in the next chapter when definingasymptotic martingales.Definition. A positive, integer valued, possibly infinite, random variable iscalled a stopping time (with respect to fFn; n 1g) if

f = ng 2 Fn; for all n 2 N:

We observe that f = ng = (f ng n f n 1g) 2 Fn and that f n 1g =

Sn1k=1f = kg: Then using the fact that Fn is a -field we get that

f ng = f = ng[f n1g 2 Fn: Furthermore f > ng = (f ng)c 2 Fn:Hence, as defined above is a stopping time if

f ng 2 Fn; for all n 2 N;

ifff > ng 2 Fn; for all n 2 N:

For variables 1; 2 on (Ω;F ;P) we use the notation 1 2 to mean that1(!) 2(!); for almost all ! 2 Ω: A stopping time is bounded iff there existsan integer M such that P ( M) = 1: We denote the set of bounded stoppingtimes by T: Now, given a sequence of random variables fXn; n 0g, we needto define some important properties it may possess with respect to the sequencefFn; n 0g:Definition. If Xn is Fn-measurable for all n then fXn; n 0g is called Fn-adapted. If Fn = fX0; X1; : : : ; Xng then fFn; n 0g is called the sequence ofnatural -algebras. Now we give a definition of a martingale.Definition. An integrable Fn-adapted sequence fXng is called a martingale ifE(Xn+1jFn) = Xn almost surely for all n 0:

We can now use the smoothing lemma to calculate the expectation of a mar-tingale: For any m;n 2 N;m < n;

E(Xn) = E(E(XnjFm)) = E(Xm):

Thus a martingale has constant expectation.Next we present some examples of martingales, most of which are found e.g

in Gut (2005), Chapter 10, where a thorough introduction of martingale theoryis given. As a first example of a martingale we may consider a one-dimensionalsymmetric random walk. Let fXn; n 0g be a sequence of random variablessuch that Xn 2 Z; is the position, after n steps, of a particle which performes asymmetric random walk on Z, i.e., the particle starts at X0 = a 2 Z; and in eachtime step moves one step up with probability 1

2 , or one step down with probability12 , so that X1 = a+ 1 with probability 1

2 and X1 = a 1 with probability 12 . Let

Fn = fX0; : : : ; Xng: Then, by independence,

E(Xn+1jFn) = E(Xn+1jX0; : : : ; Xn) = E(Xn+1jXn) = 12

((Xn+1)+(Xn1)) = Xn:

Thus fXn; n 0g is a martingale.

18

For another example we consider the sum of independent random variables.Let Y1; Y2; : : : be independent random variables with mean zero, and let Xn =Pn

i=1 Yi; n 0: Further let Fn = fY0; : : : ; Yng; which equals fX0; : : : ; Xng:Then

E(Xn+1jFn) = E(Xn + Yn+1jFn) = Xn + E(Yn+1jFn) = Xn + 0 = Xn:

Thus fXn; n 0g is a martingale.If instead Y1; Y2; : : : above have mean equal to 1, then one can define Xn =Qn

k=1 Yk; n 1; with X0 = 1: Let fFn; n 0g be the natural -algebras. ThenfXn; n 0g is a martingale.

Gut (2005) points out that the example of a martingale given above may beapplied to a gambling situation, namely double or nothing. To see this let X0 = 1and define recursively

Xn+1 =

(2Xn; with probability 1

2 ;

0; with probability 12 :

Then one may identify Xn with the productQn

k=1 Yk; n 1; with X0 = 1: whereY1; Y2; : : :, are independent random variables with mean equal to 1 (since eachof the variables is equal to 2 or equal to 0, both with probability 1

2): HencefXn; n 0g is a martingale.

8 Asymptotic martingalesMartingales are a special case of more general adapted sequences with certainconvergence properties. In this chapter a brief description shall be given of thetheory of asymptotic martingales and some preliminary theorems and lemmasshall be proved. The literature which is used in this short review includes Gut(1983), Bellow (1977), Lamb (1973) and Chacon & Sucheston (1975). Thereaftera theorem concerning the Radon–Nikodym property for vector–valued asymptoticmartingales shall be presented. This result is found in Chacon & Sucheston (1975)and Austin, Edgar & Ionescu Tulcea (1974). Finally we state and prove the maintheorem of this chapter which provides a condition for the strong convergence ofasymptotic martingales taking values in a Banach space. This result is found inBellow (1976).

8.1 Definition and brief theoryIn this section a short description of the theory asymptotic martingales, which isrequired in order to prove the main theorem of this chapter, shall be given. Webegin by giving the definition of an asymptotic martingale.Definition. If fXn; n 0g is an Fn-adapted sequence of integrable randomvariables then fXn;Fn; n 0g is called an asymptotic martingale iff limT

RX =

z exists, that is iff the net (E(X ))2T converges. This means that for every > 0there exists 0 2 T such that jEX zj < whenever > 0; 2 T:

19

Let us consider the case of an adapted sequence fXn; n 0g where E(X ) =c; 8 2 T; where c is a constant. Then it has been shown (we refer to e.g. Gut(2005), Theorem 10:7:2) that fXn; n 0g is a martingale. The converse alsoholds. To see this let m < n, Λ 2 Fm and suppose that

RX is constant for all

2 T: Define 1 = n almost surely, and

2(!) =

(m; if ! 2 Λ;n; if ! =2 Λ:

Then by assumptionR

Λ XnP +R

Λc XndP =R

Λ XmdP +R

Λc XmdP : ThusZΛXndP =

ZΛXmdP :

This shows that fXn;Fn; n 0g is a martingale iffRX is constant for all 2 T:

The following lemma shows that the positive and negative parts of real-valuedasymptotic martingales are asymptotic martingales. It shall be used to prove aconvergence theorem for certain types of asymptotic martingales.

Lemma 1. Let fXn; n 0g be a sequence of integrable random variables andsuppose supn

R jXnjdP <1: Then the following assertions are equivalent:

1) The sequence (RXdP)2T converges.

2) The sequences (RX+ dP)2T and (

RX dP)2T both converge.

Proof. We only need to show that 1) implies 2). Assume 1). Then (RXdP)2T

is bounded. Let 2 T; n and define

(!) =

((!); on fX 0g;n; on fX < 0g:

Then X+ X + jXnj so that

RX+ dP R XdP +

R jXnjdP : Then since byassumption supn

R jXnjdP < 1; this means that (R jX jdP)2T is bounded and

we have supnR jX+

n jdP <1; which in turn gives supnR jX

n jdP <1:

Next choose n0 such that for any > 0 we have that ; 2 T; n0; n0imply

(8.1) jZ

XdP Z

XdPj :

Then choose 0 2 T; 0 > n0 such that for any 2 T; 0 we have

(8.2)Z

X+ dP

ZX+0dP :

Now define

1(!) =

((!); on fX0 0g;0(!); on fX0 < 0g:

20

Then (8:1) gives j R X1dP RX0dPj : But then the relationsZ

X0dP =ZfX00g

X0dP+ZfX0<0g

X0dP =ZfX00g

X+0dP+

ZfX0<0g

X0dP ;

ZX1dP =

ZfX00g

XdP +ZfX0<0g

X0dP ;

imply that

(8.3)Z

X0dP Z

X1dP =Z

X+0dP

ZfX00g

XdP ;

where we use thatRXfX0gdP =

RX+dP ; 8X. Then (8:1) together with (8:3)

gives ZX+0dP

Zf00g

XdP + Zf00g

X+ dP +

ZX+ dP + ;

which, in turn, together with (8:2) shows that

jZ

X+ dP

ZX+0dPj ;

for any 2 T; 0: This shows that the bounded sequence (RX+ dP)2T con-

verges. Then since (RXdP)2T converges we have that (

RX dP)2T converges.

This completes the proof.

We now go on to prove a convergence theorem for real valued L1-boundedasymptotic martingales found in Austin, Edgar & Ionescu Tulcea (1974).

Theorem (real-valued asymptotic martingale) If fXn;Fn; n 0g is a real-valued L1-bounded asymptotic martingale then Xn converges almost everywhere.

Proof. By Lemma 1 we may assume, without loss of generality, that Xn 0:Now assume that the sequence Xn does not converge in order to show that inthat case Xn is not Cauchy. There are real numbers a < b such that P(A) > 0;where A = f! : lim inf Xn(!) < a < b < lim supXn(!)g: Let = (ba)P(A)

2 :We will show that given any integer M 1 there exist bounded stopping times1 M; 2 M such that

RX2dP

RX1dP (i.e. the sequence (

RX )2TdP

is not Cauchy). We prove this as follows: Let = =2b and let M 1 be anyinteger. Then there exist an integer N M and a set B 2 FN such thatP(f! : ! 2 A [ B;! =2 A \ Bg) : To see this define the index sets Ji =fk : Ck 2 Fi; Ck \ A = ;g. Then define Bi =

Sk2Ji Ck. Then fP(Bi)giM

is an increasing sequence of real numbers and P(Bi) ! P(Ac), as i ! 1.Hence there is an integer N such that jP(BN) P(A)j < =2, so that Bc

N =: Bcontains A and P(f! : ! 2 A [ B;! =2 A \ Bg) . Further there exist integersN2 > N1 > N such that if

Ω0 = f! : infNnN1

Xn(!) < a < b < supN1nN2

Xn(!)g;

21

then P(A r Ω0) : Next define C1 := f! 2 B : infNnN1 Xn(!) < ag; C2 :=f! 2 C1 : supN1nN2 Xn(!) > bg: Then

(8.4)

(C1 2 FN1 ; C2 2 FN2 ; C2 C1 B;

P(C2) P(A) 2; P(C1 r C2) 2:

Define the stopping times 1; 2 by:

1(!) =

(N1; for ! =2 C1;

inffn : N1 n N2; Xn(!) < ag; for ! 2 C1;

and

2(!) =

8><>:N1; for ! =2 C1

N2; for ! 2 C1 r C2;

inffn : N1 n N2; Xn(!) > bg; for ! 2 C2:

Then M N 1 2 and by (8:4),ZX2dP

ZX1dP =

Z(X2 X1)dP

ZC1

(X2 X1)dP

=ZC2

(X2 X1)dP +ZC1rC2

X2dP ZC1rC2

X1dP

(b a)P(C2) + 0 aP(C1 r C2) (b a)(P(A) 2) 2a = (b a)P(A) 2b= 2 = :

Thus we have shown thatRX2dP R X1dP so the sequence (

RXdP)2T

is not Cauchy, and this is true for every such sequence that does not converge.This completes the proof.

In the next section we shall consider a more general situation where the vari-ables Xn have range in a complete normed space E, i.e., Xn : Ω ! E: We nowgive some definitions that are needed to handle this case. The sequence Xn iscalled adapted if Xn : Ω ! E; is Fn-measurable in the sense of Bochner (thismeans that we use the norm on E, denoted kk, to get a real valued functionΩ ! E ! R and then decide if this function is measurable). Here we are consid-ering the probability space (Ω;F ;P), and sequences (Fn)n0 of sub--algebras ofF . An adapted sequence XN is called Bochner integrable if

R kXn(w)kdP <1.The adapted sequence is called an asymptotic martingale if limT

RXdP = z 2 E

exists, that is for every > 0 there exists 0 2 T such that kXzk < whenever > 0; 2 T ( 2 T>0): We see that the modification we need to make whencomparing to the real valued case is simply to replace the Euclidian norm by thenorm on E. Similarly we call the sequence L1-bounded if supn

R kXnkdP < 1:We now prove two lemmas and give the statement of a third lemma which willbe used in the proof of the main theorem of this chapter. The first of these issometimes called the ’maximal’ lemma.

Lemma 2. Let (Xn) be a sequence of random variables such that supTR kXkdP <

22

1: Then for each positive number a

PfsupnkXnk ag 1

asupT

ZkXkdP :

Proof. Let N be a fixed positive integer and define 2 T as follows: if n N;X1; : : : ; Xn1 < a; Xn a;X0 = 0; then set = n: If sup1nN kXnk a; set = N: Now let AN = fsup1nN kXnk > ag: Then supT

R kXkdP R kXkdP RANkXkdP aP(AN): Letting N !1 this gives Pfsupn kXnk >

ag 1a

supTR kXkdP for each positive number a. Now replacing a with a + ,

with > 0, and letting go to zero proves the statement of the lemma.

Lemma 3. Let k be a fixed positive integer, A 2 Fk: If (RXndP) is such that

limT

RXdP = z 2 E exists then (

RAXdP)2T converges.

Proof. Let N k be such that for any > 0 and 1; 1 2 TN one has j R X1dPRX1dPj < : Now let ; 2 TN ; and let N1 be an integer such that N1 >

max(; ). Then set 1 = on A, 1 = on A, 1 = 1 = N1 on Ac: Then we havef1 < Ng = f1 > N1g = ;; f1 = ng = f = ng\A 2 Fn for n 2 [N;N1); f1 =N1g = Ac 2 Fk 2 FN1 : Thus 1 is a stopping time. Analogously one gets that1 is a stopping time. Then j R

AXdP R

AXdPj = j R X1dP R X1dPj < :

This completes the proof.The next lemma shall as earlier stated only be presented and we refer to a

textbook on mathematical analysis for the proof, e.g Dunford & Schwartz (1958).The lemma is a consequence of the so called Vitali-Hahn-Saks theorem, and webelieve that it is important that the statement of the theorem itself is also given.

For this we first need to explain some notaion.Given the measure space (X;S; ) denote by v(;A) the function defined as

follows:

v(;A) = supnXi=1

j(Ai)j; A 2 S;

where the supremum is taken over all finite sequences fAig of disjoint sets in Ssuch that Ai A. We observe that if is finite then

Pni=1 j(Ai)j =

P+ (Ai) +P (Ai) = ([+(Ai))+([(Ai)) <1; whereP+;[+;

P;[ are taken overthose i such that (Ai) 0 ((Ai) < 0): Since the sequence fAig is disjoint wehave for any pair of sets, A;B 2 S; that sup

Pni=1 j(Ai)j + sup

Pni=1 j(Bi)j =

supPn

i=1 j(Ci)j; where fBig; fCig are finite sequences in B;A [ B respectively.Hence v(;A) is countably additive.Definition. A set function g on S is called -continuous if limv(;A)!0 g(A) = 0:

Now we give the statement of the Vitali–Hahn–Saks theorem followed by theimportant corollary which we shall use a a lemma in the proof of the main theoremof this paper.The Vitali–Hahn–Saks theorem. Let (X;S; ) be a measure space and letn be a sequence of -continuous, vector valued, additive set functions on S. Iflimn n(A) = (A) exists for each A 2 S; then limv(;A)!0 i(A) = 0, uniformly

23

for i=1,2,. . . .For a proof of this we refer to Dunford & Schwartz (1958), Chapter 3. The

following is the statement of a consequence of the Vitali–Hahn–Saks theorem.Corollary (Vitali–Hahn–Saks) Let n be a sequence of E-valued finite mea-sures on F (where we are considering the probability space (Ω;F ;P)) such thatlimn n(A) = (A) exists for each A 2 F : Then is a measure.Proof. For the proof of this Corollary we refer to Dunford & Schwartz (1958), p.321.

We end this section with a definition which will be of practical importance inthe formulation and proofs of the main theorems of this chapter.Definition. A complete normed space is said to have the Radon–Nikodym prop-erty if every E-valued measure on F ; of finite variation and vanishing on nullsets (with respect to P), can be represented as an integral of a random variableX in the sense that

(8.5) (A) =ZA

XdP ; A 2 F :

8.2 The Radon–Nikodym property for vector-valued asymp-totic martingales

In this section a theorem concerning the Radon–Nikodym property for vector-valued asymptotic martingales shall be proved. We refer to Chacon & Sucheston(1975).Theorem (Vector-valued asymptotic martingale). Assume that the Ba-nach space E has the Radon–Nikodym property and the dual E 0 is separable. Let(Xn)n2N be an E-valued asymptotic martingale such that

(8.6) sup2T

ZkXkdP <1

(i.e., (X )2T is L1-bounded). Then there exists an E-valued random variable Rsuch that the sequence (Xn(!))n2N converges to R(!) weakly in E, for almostevery ! 2 Ω.Proof. Let a be a positive constant and define a stopping time such that is the first n such that kXnk a; otherwise = 1 if kXnk < a for all n;Let Y = supn kXn^k: Then Y a on f = 1g: On A := f = 1g we have(by Fatou’s lemma)

RAXdP lim infn

RAkXn^kdP lim infn

R kXn^kdP :Further since the infimum of two stopping times is a stopping time we havelim infn

R kXn^kdP supTR kXkdP := M < 1: Now kXn^k kXk on

f < 1g , which implies that Y a + kX0k, and thusRY dP a + M:

Now for ; 1 2 T;R

(X^ X1^)dP =R

(Xf(^)_(^1)g Xf(1^)_(^1)g)dP ;which can be made less than any given for some 0 2 T when ( ^ 1) 2 T>0 :Hence (X^) is an asymptotic martingale. By the definition of we see that thesequence (Xn^) may only differ from (Xn) when kXk a. Then by Lemma

24

2 the measure of the set where the two sequences are different is small when ais large. Therefore we assume without loss of generality that the sequence (Xn)itself has the property

Y = supnkXnk 2 L1:

Next define a sequence of E-valued measures ; 2 T by

(A) =ZA

XdP ; A 2 F :

Let A 2 F : By Lemma 3 limT (A) = (A) exists for every A 2 Fn: Then forevery > 0 there is a set A1 2 [Fn such that P(f! : ! 2 A[A1; ! =2 A\A1g) < :Since kXk Y; 8; we have

kZA

XdP ZA1

XdPk

ZY dP :

Hence (A) = limT (A) exists for all A 2 F ; and the variation of is boundedbyRY dP . By the Corollary to the Vitali–Hahn–Saks theorem (see above) we

have that is a measure. Then the Radon–Nikodym property of E implies thatthere exists a random variable R such that

limT

ZA

XdP =ZA

RdP ; A 2 F :

Let ('i)i1 be a sequence that is dense in the unit ball of the dual space of E(i.e. the set of bounded linear functionals on E). Such a sequence exists since thedual is assumed to be separable (i.e. it has a countable dense subset). For a fixedi we have

(8.7) limT

ZA

'i(X )dP =ZA

'i(R)dP ; A 2 F :

Now we can apply the convergence theorem on real-valued asymptotic martingalesthat we have already proved to get that limn 'i(Xn) exists almost everywhere andthat therefore (8:7) implies that limn 'i(Xn) = 'i(R) except on a set of measurezero, Ωi. Since this holds for all i Xn converges weakly to R except on

Si Ωi (which

has measure zero). By the first part of the proof we have Y = supn kXnk 2 L1:Hence (Xn(!))n is finite for every ! except on a set of arbitrary small measure.This completes the proof.

8.3 Statement of the main theoremIn this section we state a theorem due to Bellow (1976) on conditions for strongconvergence of asymptotic martingales taking values in a Banach space. Ourstudies so far of the important Radon–Nikodym theorem, random variables, mar-tingales and asymptotic martingales will allow us to be able to prove this theoremin Section 8:5.Main theorem. For a Banach space E the following assertions are equivalent:

(i) E is of finite dimension.

25

(ii) Every E-valued asymptotic martingale (Xn)n2N such thatsup2T

R kXkdP <1, converges to a limit strongly almost everywhere.

(iii) Every E-valued asymptotic martingale (Xn)n2N such thatkXn(!)k 1, n 2 N and ! 2 Ω, converges to a limit strongly almosteverywhere.

8.4 Remark on a strong convergence result for martin-gales

Before we go on to prove the main theorem we wish to make a short remark on aresult by Chatterji (1968) where an elaborate proof is presented of the theoremgiven below on conditions for strong martingale convergence. We shall use thenotation of Chatterji (1968) when presenting his theorem. The main theorem onmartingale convergence by Chatterji (1968) is the following:Strong martingale convergence theorem. For a Banach space X and aprobability space (S;Σ; P ) the following statements are equivalent when holdingfor all X-valued martingales:

(1) If supn1 kfnk1 < 1; then limn!1 fn =: f1 exists strongly almost every-where.

(2) If supn1 kfnk1 < 1; exists weakly in the sense that 9f1 strongly mea-surable such that for all y 2 X 0(the dual space of X), limn!1 y(fn(s)) =y(f1(s)); for s =2 N;P(N) = 0:

(3) If supn1 jfn(s)j < C; where C > 0 is a constant, then f1 exists stronglyalmost everywhere.

(4) If for some C > 0, supn1 jfn(s)j < C, almost everywhere, then f1 existsweakly almost everywhere.

(5) If the fn are uniformly integrable (i.e. limN!1R jfnjIfjfnj>NgdP = 0; uni-

formly in n 1) there 9f1 2 L1(X;Σ) with limn!1 kfn f1k1 = 0.

(6) If supn1 kfnkp <1; 1 < p <1; then 9f1 2 Lp(X;Σ) with limn!1 kfn f1kp = 0.

(7) The space X has the RN-property with respect to the measure space (S;Σ; P ).

In particular we see that the theorem provided by Chatterji gives a resultanalogous (but stronger and at the same time restricted to Banach space val-ued martingales) to that which we have shown in the theorem on vector–valuedasymptotic martingales. This is also the reason why Chacon & Sucheston (1975)give the statement of the equivalence (1) , (7) above in their article, and alsowhy Bellow (1977) mentions this work. The major part of the proof of the theo-rem by Chatterji consists in showing (1) ) (7). One of the most important tools

26

which Chatterji uses to prove this implication is a theorem by Rickart (1943),which is a decomposition theorem for additive set functions. With this remarkmade for the case of martingales we are ready to proceed with the main theoremconcerning asymptotic martingales.

8.5 Proof of the main theorem

We will now prove the main theorem of this paper, due to Bellow (1976). Theproof that we shall give relies upon an important lemma by Dvoretzky & Rogers(1950). We shall state and prove this lemma and then also a consequence of thelemma before we give the proof of the main theorem.

The Dvoretzky–Rogers lemma. Let C be the closure of a bounded set (inan n-dimensional Euclidian space), convex with the origin as center, and let r bean integer with 1 r n: Then there are points A1; : : : ; An; on the boundary ofC such that, if 1; : : : ; r are any r real numbers with 1 r n; then the point1A1 + : : : + rAr lies in C where

(8.8) 2 =

2 + r(r 1)n

(2

1 + + 2r):

Proof. C, being convex and bounded, can be inscribed in an ellipsoid with theorigin as center having the smallest possible n-dimensional volume. We observethat once the lemma is proved for C we may perform an affine transformation ofC and the lemma will still hold. So we can assume that the ellipsoid is the unitsphere, S: We shall first show, by induction, that after an orthogonal transfor-mation has be been applied there will be r points A1; : : : ; Ar in the intersectionof C and S such that for i = 1; : : : ; r we have

(8.9)

(Ai = (ai1; : : : ; aii; 0; : : : ; 0);a2i1 + + a2

i(i1) = 1 a2ii i1

n:

Since C is inscribed in S we have for r = 1 a point A1 2 C \ S which musthave jA1j = 1 since it lies in S, so a2

11 = 1 and the statement holds. Assume thestatement for r = m 1 < n: We observe that the following ellipsoid

(8.10) (1 + )n(m1)(x21 + + x2

m1) + (1 + + 2)(m1)(x2m + + x2

n) 1;

where > 0; has m1 ’half axes’ of lengthp

(1 + )(m1)n and n (m1) ’halfaxes’ of length

p(1 + + 2)m1. This implies that the volume of the ellipsoid

is greater that the volume of the unit ball in Rn, therefore there is a pointA() = (a1; : : : ; an) on the boundary of C wich also lies in the ellipsoid given by(8:10). Since A() is on the boundary of C (and we have transformed C to theopen unit ball) it does not lie inside the unit ball hence we have a2

1 + +a2n 1:

27

Thus A() satisfies

(8.11)(1+)n(m1)(a2

1+ +a2m1)+(1++2)(m1)(a2

m+ +a2n)(a2

1+ +a2n) =

[(1 + )n(m1)1](a21 + +a2

m1) + [(1 + + 2)(m1)1](a2m + +a2

n) 0:

Now letting ! 0; through a sequence of positive real numbers we get sequenceA() which converges (all points lie in the given ellipsoid) to say Am; which by(8:10) lies in C \ S. We use Taylor expansion to see that

[(1 + )n(m1) 1] = (nm + 1) + 2 (nm + 1)(nm + 1) 12!

+ ;

[(1++2)m+1)1] = (m+1)(+2)+ ( + 2)2(m + 1)((m + 1) 1)2!

+ ;

hence as goes to zero we get from (8:11) that

(8.12) (nm + 1)(a21 + + a2

m1) + (m + 1)(a2m + + a2

n) 0:

We now recall that we have assumed existence of A1; : : : ; Am1 which satisfy(8:9). As we are considering an n-dimensional Euclidian space, we may performan orthogonal transformation of the variables x1; : : : ; xn; which leaves the pointsA1; : : : ; Am1 invariant. All of these points are independent of the last n (m1) variables xm; : : : xn so the transformation may be chosen such that the lastn (m 1) coordinates of the point Am are zero. After such a transformationwe have by (8:12) for Am

(nm+1)(a2m1+ +a2

m(m1))+(m1)a2mm = (nm+1)(1a2

mm)(m1)a2mm 0;

where we have used the induction hypothesis for the first equality. Hence

n(1 amm) m 1;

which proves (8:9) for r = 1; : : :m; and therefore completes the induction.Now let 1; : : : ; r be r real numbers with 1 r n: We use (8:9) to consider

the square of the distance of the point 1A1 + + rAr to the origin

(8.13)rX

j=1

rXi=j

iaij

!2

rX

j=1

2

0@2

ja2jj +

rX

i=j+1

iaij

!21A ;

where we have used 0 (b + c)2 =) b2 + c2 2jbcj; with b = 2ja

2jj and

c =Pr

i=j+1 iaij: Next we apply the Cauchy–Schwartz theorem to see that theright hand side of (8:13) is less than or equal to(8.14)

rXj=1

2

2ja

2jj +

rX

i=j+1

2i

! rX

k=j+1

a2kj

!! 2

rXi=1

0@a2

ii +rX

k=2

min(i1;k1)Xj=1

a2kj

1A2

i ;

where we have switched the main summation from j to i: In more detail we firstfix i = : This gives us the following terms containing the factor ,

Prk=j+1 a

2kj.

28

Here we must have j = 1; : : : 1 but also j k 1; hencerX

j=1

rX

i=j+1

2i

! rX

k=j+1

a2kj

!=

rXi=1

2i

rXk=2

Xjk1;ji1

a2kj =

rXk=2

min(i1;k1)Xj=1

a2kj

2i :

Now we use (8:9), and define the term for k = 1 to be zero, to see that

a2ii+

rXk=1

min(i1;k1)Xj=1

a2kj 1+

rXk=1

(a2k1+ +a2

k(k1)) 1+rX

k=1

k 1n

= 1+ r(r 1)n

:

Hence the right hand side of (8:14) is less that or equal to2 + r

r 1n

rXi=1

2i := 2:

Thus the point 1A1 + +rAr lies inside S and therefore belongs to C. Thiscompletes the proof of the lemma.

The lemma has the following consequence which we shall use in the proof ofthe main theorem of the paper.Corollary (Dvoretzky–Rogers) Let B be a Banach space of infinite dimensionand let c1; : : : ; cr be positive numbers. Then there exist points x1; : : : ; xr in B suchthat kxik2 = ci; i = 1; : : : r and such that

(8.15) kX0

xik2 3X0

ci;

whereP0 denotes summation over any subset of the numbers 1; : : : ; r:

Proof. Choose n = r(r1): Since B is of infinite dimension there exist n linearlyindependent elements z1; : : : ; zn: Then C = f(u1; : : : ; un) : ku1z1+ +unznk 1gis a closed convex set with the origin as center in n-dimensional Euclidian space.Let A1; : : : ; Ar be the points provided by the Dvoretzky–Rogers lemma, and set

xi = c1=2i (ai1z1 + + ainzn); i = 1; : : : ; r:

Since A1; : : : ; Ar are on the boundary of C, we have kxik2 = ci; i = 1; : : : ; r:Furthermore the point

P0 c1=2i Ai lies in C where 2 =

2 + r(r1)

r(r1)

P0 ci; thus(8:15) is satisfied. This completes the proof.

We are now ready to give the proof of the main theorem for convergence ofasymptotic martingales in finite dimensional Banach spaces.Proof of the main theorem.(i) =) (ii): Let fu1; : : : ; ung be a basis for E: Then for each n 2 N; ! 2 Ω wecan write

Xn(!) = X1n(!)u1 + + Xr

n(!)ur;

which for each 1 j r gives us a real-valued asymptotic martingale, (Xjn)nN,

with supn2NR jXj

njdP < 1: Hence strong convergence almost everywhere ofthe sequence (Xn)n2N is equivalent to coordinatewise convergence. This in turn

29

follows from the theorem on real–valued asymptotic martingales which we havealready proved.(ii) =) (iii): Since kXn(!)k 1 for all n 2 N; ! 2 Ω =) sup2T

R kXkdP <1.(iii) =) (i): We shall suppose that E has infinite dimension and then constructan E-valued asymptotic martingale with kXn(!)k 1 that does not convergestrongly almost everywhere. Define for each integer n, fAn

i g; 1 i 2n by

Ani =

hi 12n

;i

2n:

We now apply the corollary of the Dvoretzky–Rogers lemma to (c1=2i )(n) := dni =

12n ; 1 i 2n; and get n vectors xni = eni d

ni ; where the eni are distinct unit

vectors, such that

(8.16) kX0

eni dni k2 3

X0(dni )2:

Now define

Xn(!) =2nXi=1

IAni(!)eni ; ! 2 Ω;

where IAni

is the indicator function for Ani ; so that F = (X1; : : : ; Xn) is the

-algebra spanned by fAni g; 1 i 2n: We will now show that (Xn)n2N is an

asymptotic martingale. To see this let A 2 FN ; N 2 N and take 2 T; such that N: Then there is an integer K such that K N: For each N n Kwe have with the following definition f = ng \A = [j2Bn

Ani = Ωn (where Bn is

the index set corresponding to the summationP0 over any subset of the numbers

1; : : : ; r), thatZA

XdP =KX

n=N

Zf=ng\A

Xj2Bn

IAnjenj

!dP =

KXn=N

Xj2Bn

P (Ani )eni

!:

By (8:16) we have for each N n K that

kXj2Bn

P (Aj)enj k p

3

Xj2Bn

P (Aj)2

!1=2

p

3

12nXj2Bn

P (Aj)

!1=2

=p

3(p

2)n(P (Ωn)1=2

p3

(p

2)n:

Hence we have

kZA

Xk dP KX

n=N

kXj

0P (Aj)enj k

p3

KX

n=N

1p2n

! ;

since N can be chosen so large that for any > 0 we haveP

nNp

3(p

2)n : Wehave thus proved that

A 2 [n2NFn =)ZA

XdP ! 0;

30

which shows that (Xn)n2N is an asymptotic martingale. In particular we have

A 2 [n2NFn =)ZA

XndP ! 0:

Furthermore we have kXn(!)k = 1; n 2 N; ! 2 Ω; thus

(8.17)ZA

XndP ! 0; for all A 2 F : = ([n2NFn):

We recall that we are working under the assumption that E has infinite dimension.Now suppose that (Xn)n2N converges strongly almost everywhere, ! 2 Ω, that is:limn2N Xn(!) = X(!): Then by the Lebesgue dominated convergence theoremwe get Z

A

Xn(!)dP =ZA

X(!)dP ; for every A 2 F :

This together with (8:17) impliesZA

X(!)dP = 0; for all A 2 F ;

and since X is F -measurable we get X(!) = 0 for almost every ! 2 Ω: But thiscontradicts the assumption of strong convergence: limn2N Xn(!) = X(!); sincekXn(!)k = 1; n 2 N; ! 2 Ω: Hence under the assumption that E has infinitedimension, the assumption that (Xn)n2N converges strongly almost everywhere,! 2 Ω, does not hold.This completes the proof of the theorem.

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