Some applications of the Radon-Nikodymtheorem to asymptotic martingales

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

Examensarbete i matematisk statistik, 20 pongHandledare och examinator: Allan GutMars 2006

Department of MathematicsUppsala University

Some applications ofthe RadonNikodym theoremto asymptotic martingalesSupervisor Professor Allan GutAuthor Abtin DaghighiAbstractThe RadonNikodym 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 RadonNikodym 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.

1

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.

2

Contents1 Introduction

4

2 Preliminaries

4

3 Statement of the RadonNikodym theorem

6

4 Some results required for the proofs of the RadonNikodym theorem65 Two different proofs of the RadonNikodym theorem5.1 A first proof of the RadonNikodym theorem . . . . . . . . . . . .5.2 A second proof of the RadonNikodym theorem . . . . . . . . . .

131314

6 Conditional probability

15

7 Martingales

17

8 Asymptotic martingales8.1 Definition and brief theory . . . . . . . . . . . . . . . . . . . . . .8.2 The RadonNikodym property for vector-valued asymptotic martingales . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .8.3 Statement of the main theorem . . . . . . . . . . . . . . . . . . .8.4 Remark on a strong convergence result for martingales . . . . . .8.5 Proof of the main theorem . . . . . . . . . . . . . . . . . . . . . .

1919

9 References

31

3

24252627

1

Introduction

In this paper we review some important fields in probability theory, namely conditional probability, martingales and asymptotic martingales. We shall then investigate the use of the RadonNikodym theorem in the theory of asymptoticmartingale convergence.The RadonNikodym theorem provides conditions for existence of a certain integral representation of measures. The paper first presents and proves the RadonNikodym theorem. This theorem was first proved for Rn by Johann Radon (18471956) 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 literature used for the presentation of the RadonNikodym 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 conditional probability. Then some basic martingale theory will be presented. Thetheory of asymptotic martingales shall be reviewed briefly and an application ofthe RadonNikodym 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 RadonNikodymproperty 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 dimension. 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

Preliminaries

The reader is assumed to have taken an introductory course in measure theory.This means that we shall in the text refer to some wellknown theorems withoutproof. These include: The monotone convergence theorem, Fatous 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 RadonNikodym 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 a4

measure, , on S . A set E is called positive (negative) if 8F such thatmeasurable E \ F is also measurable and (E \ F ) 0 ((E \ F ) 0).

F is

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 onePof the values +1 and 1 (inparticular, if fEn gn is a disjoint sequence then 1n=1 (En ) is either convergentor 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:

P

Definition. A simple function f = ni=1 ai IEi (x); where thecalled integrable if (Ei ) < 1 for all i for which ai 6= 0.

Ei are disjoint, is

Definition. A nonnegative function f is called S -measurable if there exists anincreasing sequence, fsn gn , 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 ffn g of integrableR simple functions which convergein measure to f . The integral of f , denoted fd, is defined by

Z

fd = nlim!1

Z

fn d:

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 ffn g be an increasing sequence ofextended real valued nonnegative measurable functions such that limn!1 fn (x) =f (x) almostThen f is a nonnegative measurable function andR everywhere.Rlimn!1 fn d = fd:Fatous lemma.then

If

ffng is a sequence of nonnegative measurable functions,

Z

lim inf

n!1

fn d lim ninf!1

Z

fn d:

The Lebesgue dominated convergence theorem. If ffn g is a sequence ofintegrable functions which converges in measure () to f , and if g > 0 is anintegrable function such Rthat jfn (xR)j g (x) almost everywere, n = 1; 2; : : : ; thenf is integrable and limn fn d = fd:In probability theory the set we have often referred to above as X is denotedby and F denotes a -algebra (of subsets on ):Definition. A probability measure, denotedwith domain F such that:5

P ; on F

is a real valued function

(i)(ii)(iii)

8e 2 F ; P (e) 0;P () = 1:SPIf fen g is a countable family of disjoint sets in F then P ( n en ) = n P (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 (; 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 : X (!) bg 2 F :

3

Statement of the RadonNikodym theorem

The theorem known as the RadonNikodym theorem for signed measures gives arepresentation involving a measurable function for signed measures.The RadonNikodym 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 realvalued measurable function f on X such that either f +or f is integrable and

(E ) : =

Z

fd =

Z

f d +

Z

f dRfor every measurable set E . Furthermore, if (E ) = E gd, E 2 S; then fx 2X : 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 .d , and called the RadonNikodymThe function f is sometimes denoted dE

+

E

E

derivative.

4

Some results required for the proofs of theRadonNikodym theorem

In this chapter we present some of the theorems and propositions whose resultswill be used in the proofs of the RadonNikodym 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 respectively 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 .6

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 measurableS1 negative sets such that limi!1 (bi ) = . Bythe argument above B = i=1 bi is a measurable negative set and furthermoresuch that (B ) is minimal.Now consider the set X r B and assume that this set is not positive in orderto arrive at a contradiction.Let E0 be a measurable subset of X r B such that (E0 ) < 0 (such a setexists 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 ) k11 ((E1 ) will be finite since (E0 ) < 0). Now since

(E0 r E1 ) = (E0 ) (E1 ) (E0 )

1

k1

< 0;

the argument which has been applied to E0 may be applied to E0 r E1 also. Thuschoosing k2 as the smallest positive integer such that there is a measurable setE2 E0 r E1 with the property (E2 ) k12 , and so on. This describes a procedureSthat can be iterated to infinity. Furthermore Ej +1 E0 n ji=1 Ei , thus the Eiare disjoint.Now for any measurable subset G E0 , j(G)j < 1: This is true sincej(E0)j < 1, and (E0) = (E0 r G) + (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 r G) and (G) are both finite.SThis means that limn!1 k1n = 0, since we can set G = 1j =1PEj ,1 which is acountable union of measurable subsets of E0 , and 1 > (G) = n kn : Define

F0 = E0 r

1[

j =1

Ej :

Then, by construction, F0 must be a negative set since if there existed a positivesubsetS E such that (F0 r E ) (E ) then Ei would already have been includedin 1j =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 r B is a positive set and by choosing A = X r B 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.7

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 RadonNikodymtheorem 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 wenotation Lp (X; B ; ) for the set of S -measurable functions, f , such thatRuse thepjf j d < 1). Then there exists a unique g0 2 L2(X; S; ) such that

T (f ) = hf; g0 i 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; g0 i, 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

2 KerT , then g ? g. We will find the desired g0 as follows: Define g0 :=(g )g: For all z 2 C; T (zg) = T (g)z and hzg; g0 i = z hg; g0 i = z hg; hTg;gi gi =

since if gT (g )hg;gi

T (g)z: Hence(4.1)Also, if

f

2 Ker(T ); then

T (zg) = hzg; g0 i; 8z 2 C:

(4.2)and if

f 2= Ker(T ); then (f

T (f ) = 0 = hf; g0 i;

zg) 2 Ker(T ) iff

0 = T (f

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

f 2= Ker(T ); it follows thatwhich gives zand therefore, by (4:1) and (4:2); that= TT ((fg)) : Hence for

T (f ) = T f

T (f )T (f )g+T g= hfT (g)T (g)8

g

f

g TT ((fg))

2 Ker(T )

T (f )T (f ); g0 i + hg; g i = hf; g0 iT (g)T (g) 0

(where (4:1) has been applied to the term

T

g TT ((fg))

and (4:2) has been applied

T f g TT ((fg)) ). This proves existence.Now, if hf; g0 i = hf; g1 i 8f 2 L2 (X; S); then hf; g0 g1 i = 0 8f 2 L2 (X; S):In particular, letting f = g0 g1 gives 0 = kg0 g1 k, so that g0 = g1 ; which provesto the term

uniqueness. This completes the proof.

Proposition 1. Let f 2 L1 (X; S;); < 1; Y C; Y be closed. If for anyRE 2 S such that (E ) > 0 we have (1E ) E fd 2 Y; then f (x) 2 Y for almostevery x 2 X with respect to .

Proof. It suffices to show that (fx : f (x) 2= Y g) = 0: Since C rSY = Y c is open,cthere exist countably many open balls fBn gn1 such that Y = 1n=1 Bn : Thus,

fx 2 X : f (x) 2= Y g = fx 2 X : f (x) 2

1[

n=1

Bn g =

1[

n=1

f 1 (Bn ):

But f is measurable and we have f 1 (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 = f 1 (Bn ) and suppose that (En ) > 0; in order to arriveat a contradiction. ThenZZ11j (E ) fd z0j = j (E ) (f (x) z0)djn Enn EnZ1 (E ) jf (x) z0jd r0;n En

R

which implies that (E1 n ) En fd 2 Bn Y c ; which is a contradiction to theRassumption (1E ) E fd 2 Y: Hence (En ) = 0; which proves the statement.The von Neumann theorem. Let ; be two -finite measures on a measurable space (X; S ). Then there exist mutually disjoint sets Xi 2 S; 1 i 3; suchthat the following hold:(i)(ii)

X=

S3

i=1 Xi ;

(X3 ) = (X1 ) = 0;

(iii) There exists a nonnegative measurable function g on X such that g (x)0 8x 2 X2 ; and such that for all E 2 S with E X2 we have

(E ) =

Z

E

>

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 ( + )9

and by the CauchySchwarz inequality,

Z

jf jd( + ) (( + )(X ))

1=2

Now, if f 2 L1 ( + ) thenT (f ) : L2 ( + ) ! R as

R

Z

1=2

jf jd( + )

R

jf jd jf jd( + ) which gives f 2 L1( ): Define

T (f ) =

Z

fd; f

2 L2( + ):

Then T is a bounded linear functional since jT (f )jRiesz representationtheorem there exists a functionRT (f ) = ff0 d( + ); 8f 2 L2 ( + ); so that

Z

Putting(4.3)

:

fd =

Z

( (X0))1=2kf k2: By thef0 2 L2 ( + ); such that

ff0 d( + ):

f = IE ; E 2 S where IE is the indicator function for E , givesZ (E ) = f0 d( + );

for every E 2 S:Using Proposition 1 with

E

Y = R0 ; f = f0 ; and replaced by ( + ) gives

( + )fx 2 X :

(4.4)

f0 (x) 2= R0 g = 0:

Thus f0 is real-valued (so f0 = f0 ) for almost every x 2 X and fR0 (x) 0 foralmost every x with respect to ( +R ): Also ( + )(E ) (E ) = E f0 d( + );for all E 2 S; which implies that E (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;X10 := fx 2 X : f0 (x) = 1g; X1 = X10 [ N;X2 := fx 2 X : 0 < f0 (x) < 1g;X3 := fx 2 X : f0 (x) = 0g:S3Then ( + )(N ) = 0 andR Xi ; i = 1; 2; 3 are pairwise disjoint with X = i=1 Xi :Furthermore, (X3 ) = X3 f0 d( + ) = 0 and

(X 0 ) = ( + )(X 0 )1

1

(X 0 ) =1

Z

X10

(N ) = 0 thenR implies R (X1 ) = 0. Also (E ) =that (E )E f0 d = E f0 d: ThusZZ(1 f0 )d =f0 d;E

Therefore,

E

ZX2

s(1 f0 )d =10

ZX2

(1

R

f0 )d( + ) = 0:

E f0 d( + ); for all

8E 2 X2:sf0 d;

E X2 ; so

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 fsi gi we get

Z

lim

i!1 X2

For

si (1 f0 )d =

E X2 ; define

(

f (x) :=Then

Z

1

f (1 f0 )d =

IE (x)f0 ( x ) ;

0;

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

g(x) :=

X2

1

R

Z

X2

ff0 d:

x 2= X1 ;otherwise.

E

1

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

f0 ( x )f0 ( x ) ;

0;

x 2 X2 ;otherwise.

Then g is a nonnegative measurable function and, 8E

(E ) =

Z

E

2 S with E X2; we have

gd:

This proves the theorem for the case of finite measures.SFor the case where ; are -finite measures one can write X = 1i=1 Yi ; wherethe sets Yi 2 S are pairwise disjoint, (Yi ) < 1 and with (Yi ) < 1 8i 1: Bythe proof of the finite case above one can find, for every i 1; mutually disjointsets Xi1 ; Xi2 ; Xi3 2 S; such that Yi = Xi1 [ Xi2 [ Xi3 with (Xi1 ) = (Xi3 ) = 0:Also, for every i 0; there exists a measurable function gi such that gi (x) > 0for almost everyx 2 Xi2 and gi = 0 for almost every x 2 Xi1 [ Xi3 : FurthermoreR (E \ Xi2 ) = (E \Xi2 ) gi d; for all E 2 S: Now defineSS1 2S1 33X1 := 1i=1 Xi ; X2 := i=1 Xi ; X3 := i=1 Xi ; and

(

g: =Then X1 ; X2 ; X3 andthe proof.

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

g satisfy the requirements of the theorem. This completes

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

a (E ) =for every

E 2 S:11

Z

E

fd;

f such that

(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 X2c , such that

(E \ X2 ) =

Z

E \X2

gd; 8E 2 S:

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

(

f :=Then

g(x); x 2 X2 ;0;x 2 X1 [ X3 :

f is a nonnegative measurable function on X andZa (E ) = fd;8E 2 S:E

This proves statements (i) and (ii): It remains to prove uniqueness. Assume thereexist measures a0 and that s0 ; a set A0 and a nonnegativefunction f 0R measurable0000c000such that = a + s ; ((A ) ) = s (A ) = 0; a (E ) = E f d; 8E 2 S:Then((A0 \ A)c ) = (A0c [ Ac ) = 0 = s (A0 \ A) = s0 (A0 \ A):Furthermore,(4.5)

a ((A0 \ A)c ) = a0 ((A0 \ A)c ) = 0;

since the measure of the given set with respect to is zero. Since a (E )+ s (E ) =a0 (E ) + s0 (E ); it follows that(4.6)

a (E ) a0 (E ) = s (E ) s0 (E );

8E 2 S such that (E ) < +1: Since f

is nonnegative and

subset of a set of measure zero we get(4.7)

E \ (A0 \ A)c is a

a (E \ (A0 \ A)c ) a0 (E \ (A0 \ A)c ) = 0:

We now note that a (E ) = a (E \ (A0 \ A)c ) + a (E \ (A0 \ A)); and a0 (E ) =a (E \ (A0 \ A)c ) + a0 (E \ (A0 \ A)): Making this substitution in (4:6) and thenapplying (4:5) gives

a (E \ (A0 \ A)) a0 (E \ (A0 \ A)) = s (E \ (A0 \ A)) s0 (E \ (A0 \ A)) = 0:

Thus

8E 2

a (E \ (A0 \ A)) = a0 (E \ (A0 \ A));S with (E ) < +1; which together with (4:5) shows that a (E ) =12

a0 (E ); 8E 2 S such that (E ) < +1: Since is -finite this means thata (E ) = a0 (E ): An analogous argument for s gives s (E ) = s0 (E ): This completes the proof.

The second proof of the RadonNikodym 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. Letmeasure

X = An [ Bn be a Hahn decomposition with respect to the signed

n ; n = 1; 2; : : :, and let11[\A0 = An ; B0 = Bn :n=1

n=1

B0 Bn then gives 0 (B0 ) n1 (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 = N1 so that therequirements of the theorem are satisfied. This completes the proof.

5

Two different proofs of the RadonNikodymtheorem

This chapter will present two different proofs of the RadonNikodym 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 RadonNikodym theorem

In this section we shall use important theorems from the preceding chapter togive a proof of the RadonNikodym theorem (we refer to Rana (1997)):Proof of the RadonNikodym 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 theR same theorem there is a nonnegative measurable function f such thata (E ) = E fd; for every E 2 S: To prove uniqueness assume existence of g 6= fas 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: Then

Z

A\E

(f (x)

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

which is a contradiction. Thus f (x) g (x) for almost everythe theorem for the case of -finite measures.13

x 2 X . This proves

Now let

be signed. We shall next use the following statement:()

+ and

iff

;

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 RadonNikodym theorem

This 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 RadonNikodym theorem for signed measures.Consider the case where both and are finite measures. Let K be the classofR all nonnegative functions, f , that are integrable with respect to , such thatE fd (E ) for every measurable set E . Furthermore let

= supLet

(Z

)

fd : f

2K :

ffng be a sequence of functions in K such thatZ

lim

n!1

fn d = :

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 gives

Z

E

gn d

n ZXj =1

fj d

nXj =1

(Ej ) = (E ) < 1;

since we have assumed ; to be finite. This shows that fgn gn is an increasing sequence 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 measurableand since we have assumed ; to be finite f0 is also integrable.RR FurtherE limn!1 gn d (E ) for every measurable set E gives f0 2 K and f0 d = :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

Z

0 (E ) = (E )

E

fd:

This measure must be identically zero. To prove this assume that it is not identically 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)

Z

E \A

fd;

E . If g = f + IA , thenZfd + (E \ A) fd + (E \ A) (E );

for every measurable set

Z

E

gd =

Z

E

E rA c

for every measurable set

E . Thus g 2 K. ButZZgd = fd + (A) > ;E

E

which is a contradiction to the maximality of f . Hence the assumption that 0 isnot identically zero does not hold.RThis thus proves the existence of f such that (E ) = E fd, for the case offinite and . Now f is integrable since is finite. AssumeR that there exists f1which also fulfills the requirements of the theorem. Then F (fR f1 )d = 0; forevery measurable set F: Let E = fx : f (x) f1 (x) > 0g, then E (f f1 )d = 0.This implies that (E ) = 0. To see that this is true, let En = fx : (f (x) f1 (x)) 1n g, n = 1; 2; : : :. Now

Z

f1 )d (E \ En ) 0:nE \ EnS1PBut since E = n=1 En , the relation (E ) 1n=1 (En \ E ) implies (E ) = 0.Replacing (f f1 ) with (f1 f ) shows that fx : (f (x) f1 (x)) < 0g is also a set0=

1

(f

of measure zero. This proves uniqueness.Finally, X is a countable, disjoint union of measurable sets Ek ; k 2 AP, where Ais a countable index set, on which ; are finite. Therefore writingf=RP kR2A fk Ikwhere Ik denotes the indicator function for Ek one gets that f = k fk d:Using the statement () from the first proof we have()

R

+ and :RR+Now we can write (E ) := E fd = E f + dE f d = + and since iff

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

6

Conditional probability

In this chapter an application of the RadonNikodym theorem to probabilitytheory shall be presented. The result is found in, e.g., Gut (2005) or Chung15

(1968) and concerns the definition of conditional probability. We begin with adefinition.Definition. If is a set in F such thatprobability relative to as follows

P () > 0; then define the conditional

\ E) :P(E ) = P (P ()Further, the conditional expectation relative to is

Z

1E(Y ) = Y (!)P(!) = P ()

Z

Y (!)P (!):

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

E (Y jG ) =

Xn

E (Y )I ;n

n

where In is the indicator function. This variable takes the valueset n : Thus

E (Y ) =

1 ZX

E (Y ) on then

1X

Y (!)P (!) =P (n)En (Y )n=1Z1 ZXE (Y jG )dP = E (Y jG )dP :=

n=1

n

n=1

n

S

Further, any 2 G is the union of a subcollection of the n , say j 2J N j ;RRP RP Rwhich gives Y dP = j j Y dP = j j E (Y jG )dP = E (Y jG )dP : Thus

8 2 G :

(6.1)

Z

Y dP =

Z

E (Y jG )dP :

'1 ; '2 such thatZE (Y jG )dP = E 'idP ; i = 1; 2:

Now assume that there exist two functions

8 2 G :

Z

Y dP =

Z

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 variables16

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

Z

8 2 G : Y dP = E (Y jG )dP :

(6.2)

Proof. Consider the set function

on G :

Z

8 2 G : () = Y dP :

It is finite valued and countably additive, hence a signed measure on G : If P () =0 then () = 0; hence P . Then by the RadonNikodym theorem thetheorem is proved and we have that E (Y jG ) coincides with the derivative ddP :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 such that(i) it belongs to G ,(ii) it has the same integral as

7

Y over any set in G .

Martingales

In 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 RadonNikodym 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 (; 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.

F1 F2; thenE (E (X jF2)jF1) = E (X jF1) = E (E (X jF1)jF2);

The smoothing lemma. If

almost surely.

Proof. Let 2 F1

Z

F2: Then

Z

Z

Z

E (E (X jF2)jF1)dP = E (X jF2)dP = XdP = E (X jF1)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 (X jF1 ) 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 variablecalled a stopping time (with respect to fFn ; n 1g) if

f = ng 2 Fn;

for all

is

n 2 N:

We observeSn 1that f = ng = (f ng n f n 1g) 2 Fn and that f n 1g = k=1 f = kg: Then using the fact that Fn is a -field we get thatf ng = f = ng[f n 1g 2 Fn: Furthermore f > ng = (f ng)c 2 Fn:Hence, as defined above is a stopping time if

iff

f ng 2 Fn;

for all

n 2 N;

f > ng 2 Fn;

for all

n 2 N:

For variables 1 ; 2 on (; F ; P ) we use the notation 1 2 to mean that1 (!) 2 (!); for almost all ! 2 : 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 ; : : : ; Xn g then fFn ; n 0g is called the sequence ofnatural -algebras. Now we give a definition of a martingale.Definition. An integrable Fn -adapted sequencealmost surely for all n 0:

E (Xn+1jFn) = Xn

fXng is called a martingale if

We can now use the smoothing lemma to calculate the expectation of a martingale: 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.gin 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 12 , or one step down with probability1, so that X1 = a + 1 with probability 12 and X1 = a 1 with probability 12 . Let2Fn = fX0; : : : ; Xng: Then, by independence,

E (Xn+1jFn) = E (Xn+1jX0; : : : ; Xn) = E (Xn+1jXn) = 12 ((Xn +1)+(XnThus fXn ; n 0g is a martingale.18

1)) = Xn :

For another example we consider the sum of independent random variables.Let Y1 ; Y2 ; : : : be independent random variables with mean zero, and let Xn =Pni=1 Yi ; n 0: Further let Fn = fY0 ; : : : ; Yn g; which equals fX0 ; : : : ; Xn g:Then

E (Xn+1jFn) = E (Xn + Yn+1jFn) = Xn + E (Yn+1jFn) = Xn + 0 = Xn:Thus fXn ; n 0g is a martingale.

Qn If instead Y1 ; Y2 ; : : : above have mean equal to 1, then one can define Xn =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 ;0;

with probabilitywith probability

Q

1212

;:

Then one may identify Xn with the product nk=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 21 ): HencefXn; n 0g is a martingale.

8

Asymptotic martingales

Martingales 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 RadonNikodym property for vectorvalued 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 theory

In 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 randomRvariables then fXn ; Fn ; n 0g is called an asymptotic martingale iff limT X =z exists, that is iff the net (E (X )) 2T converges. This means that for every > 0there exists 0 2 T such that jEX z j < 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.R The converse alsoholds. To see this let m < n, 2 Fm and suppose that X is constant for all 2 T: Define 1 = n almost surely, and

(

m; if ! 2 ;n; if ! 2= :RRRRThen by assumption Xn P + c Xn dP = Xm dP + c Xm dP : ThusZZXn dP = Xm dP :RThis shows that fXn ; Fn ; n 0g is a martingale iff X is constant for all 2 (!) =

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. RLet fXn ; n 0g be a sequence of integrable random variables andsuppose supn jXn jdP < 1: Then the following assertions are equivalent:

R

X dP ) 2T converges.RR2) The sequences ( X+ dP ) 2T and ( X dP ) 2T both converge.1) The sequence (

R

Proof. We only need to show that 1) implies 2). Assume 1). Then (is bounded. Let 2 T; n and define

(

X dP ) 2T

(!); on fX 0g;n; on fX < 0g:R +RRThen X+ X +R jXn j so that X dP X dPR + jXn jdP : Then since byassumption supR n jXn jdP < 1; this means that ( jXR jdP ) 2T is bounded andwe have supn jXn+ jdP < 1; which in turn gives supn jXn jdP < 1:Next choose n0 such that for any > 0 we have that ; 2 T; n0 ; n0(!) =

imply(8.1)

Then choose(8.2)

Z

j X dP

Z

X dPj :

0 2 T; 0 > n0 such that for any 2 T; 0 we haveZZ+X dPX+0 dP :

Now define

(

1 (!) =

(!); on fX0 0g;0 (!); on fX0 < 0g:20

Then (8:1) gives

Z

X0 dP =

Z

R

Z

Z

(8.3)

X0 dP

Z

fX0 N such that if 0 = f! :

inf

N nN1

Xn (!) < a < b < sup Xn (!)g;N1 nN2

21

P (A r 0) : Next define C1 := f! 2 B : inf N nN Xn(!) < ag; C2 :=f! 2 C1 : supN nN Xn(!) > bg: Then(C1 2 FN ; C2 2 FN ; C2 C1 B;(8.4)P (C2) P (A) 2; P (C1 r C2) 2:

then

1

1

2

1

Define the stopping times

(

1 (!) =

1 ; 2 by:

! 2= C1 ;inf fn : N1 n N2 ; Xn (! ) < ag; for ! 2 C1 ;N1 ;

for

8>

:inf fn : N

and

! 2= C1! 2 C1 r C2 ;for ! 2 C2 :1 n N2 ; Xn (! ) > bg;

Then

Z

2

M

forfor

N 1 2 and by (8:4),

X2 dP

Z

X1 dP =

Z

(X2

Z

=

C2

X1 )dP

(X2

Z

X1 )dP +

ZC1

(X2

C1 rC2

X1 )dPZX2 dP

C1 rC2

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

R

= :

R

X1 dP

2b

R

Thus we have shown that X2 dPX1 dP so the sequence ( X dP ) 2Tis 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 variables Xn have range in a complete normed space E , i.e., Xn : ! E: We nowgive some definitions that are needed to handle this case. The sequence Xn iscalled adapted if Xn : ! E; is Fn -measurable in the sense of Bochner (thismeans that we use the norm on E , denoted kk, to get a real valued function ! E ! R and then decide if this function is measurable). Here we are considering the probability space (; F ; P ), and sequences (Fn )n0 Rof sub- -algebras ofF . An adapted sequence XN is called Bochner integrable if kRXn(w)kdP < 1.The adapted sequence is called an asymptotic martingale if limT X dP = z 2 Eexists, that is for every > 0 there exists 0 2 T such that kX z k < 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 EuclidianR norm by the1norm on E . Similarly we call the sequence L -bounded if supn kXn kdP < 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 supT22

R

kX kdP ag: Then supT kX kdP kX kdP R AN kX kdP aP (AN ): Letting N ! 1 this gives Pfsupn kXnk >ag a1 supT kX kdP for each positive number a. Now replacing a with a + ,with > 0, and letting go to zero proves the statement of the lemma.RLemma3.Letkbeafixedpositiveinteger,A2F:If(Xn dP ) is such thatkRRlimT X dP = z 2 E exists then ( A X dP ) 2T converges.RProof.Let N k be such that for any > 0 and 1 ; 1 2 TN one has j X1 dPRX1 dPj < : 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 < N g = f1 > N1g = ;; f1 = ng = f = ng\ A 2 Fn for n 2 [N; N1); f1 =N1 g = Ac 2 Fk 2 FN1 : ThusR1 is a stoppingtime. AnalogouslyRRone gets thatRX1 dPj < :1 is a stopping time. Then j A X dP A X dPj = j X1 dPProof. Let

This completes the proof.

The next lemma shall as earlier stated only be presented and we refer to atextbook 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 asfollows:

v(; A) = sup

nXi=1

j(Ai)j; A 2 S;

where the supremum is taken over all finite sequencesPfnAi g of disjointP+sets in SsuchthatA

A.Weobservethatif

isfinitethenj

(A)j=(Ai ) +iPP+ +i=1P i+(Ai ) = ([ (Ai )) + ([ (Ai )) < 1; where;[ ;; [ are taken overthose i such that (Ai ) 0 ((Ai ) < 0): SincePthe sequence fAi gPis disjoint wehavePfor any pair of sets, A; B 2 S; that sup ni=1 j(Ai )j + sup ni=1 j(Bi )j =sup ni=1 j(Ci )j; where fBi g; fCi g 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 VitaliHahnSaks theorem followed by theimportant corollary which we shall use a a lemma in the proof of the main theoremof this paper.The VitaliHahnSaks 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, uniformly23

for i=1,2,. . . .For a proof of this we refer to Dunford & Schwartz (1958), Chapter 3. Thefollowing is the statement of a consequence of the VitaliHahnSaks theorem.Corollary (VitaliHahnSaks) Let n be a sequence of E -valued finite measures on F (where we are considering the probability space (; 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 RadonNikodym property 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)

8.2

(A) =

Z

A

XdP ;

A 2 F:

The RadonNikodym property for vector-valued asymptotic martingales

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

Z

(8.6)

sup 2T

kX kdP < 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 .

Proof. Let a be a positive constant and define a stopping time such that is the first n such that kXn k a; otherwise = 1 if kXn k < a for all n;Let Y = supn kXn^ kR: Then Y a on fR = 1g: On A := f = R1g we have(by Fatous lemma) A X dP lim inf n A kXn^ kdP lim inf n kXn^ kdP :Further Rsince the infimum ofR two stopping times is a stopping time we havelim inf n kXn^ kdP supT kX kdP := M < 1: Now kXR n^ k kX k onf < 1g , whichR implies that Y a +R kX0k, and thus Y dP a + M:Now for ; 1 2 T; (X ^ X1 ^ )dP = (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 kX k a. Then by Lemma24

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 propertyY = sup kXn k 2 L1 :n

E -valued measures ; 2 T byZ (A) = X dP ;A 2 F:

Next define a sequence of

A

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 \ A1 g) < :Since kX k Y; 8; we have

Z

k X dPA

Z

A1

X dPk

Z

Y dP :

HenceR (A) = limT (A) exists for all A 2 F ; and the variation of is boundedby Y dP . By the Corollary to the VitaliHahnSaks theorem (see above) wehave that is a measure. Then the RadonNikodym property of E implies thatthere exists a random variable R such that

Z

limT

A

X dP =

Z

A

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

Z

(8.7)

limT

A

'i (X )dP =

Z

A

'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 setSof measurezero, i . Since this holds for all i Xn converges weakly to R except on i i (whichhas measure zero). By the first part of the proof we have Y = supn kXn k 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 theorem

In 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 RadonNikodym theorem, random variables, martingales and asymptotic martingales will allow us to be able to prove this theoremin Section 8:5.Main theorem. For a Banach space(i)

E the following assertions are equivalent:

E is of finite dimension.25

(ii) Every ER -valued asymptotic martingale (Xn )n2N such thatsup 2T kX kdP < 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 , converges to a limit strongly almosteverywhere.

8.4

Remark on a strong convergence result for martingales

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; ; P ) the following statements are equivalent when holdingfor all X -valued martingales:(1) If supn1 kfn k1where.

0 is a constant, then f1 exists strongly

(4) If for some C > 0, supn1 jfn (s)jweakly almost everywhere.

< C , almost everywhere, then f1 existsR

(5) If the fn are uniformly integrable (i.e. limN !1 jfn jIfjfn j>N g dP = 0; uniformly in n 1) there 9f1 2 L1 (X; ) with limn!1 kfn f1 k1 = 0.(6) If supn1 kfn kpf1 kp = 0.

< 1; 1 < p < 1; then 9f1 2 Lp (X; ) with limn!1 kfn

(7) The space X has the RN-property with respect to the measure space (S; ; 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 valued martingales) to that which we have shown in the theorem on vectorvaluedasymptotic 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 theorem by Chatterji consists in showing (1) ) (7). One of the most important tools26

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 DvoretzkyRogers 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 point1 A1 + : : : + r Ar lies in C where

= 2+2

(8.8)

r(r

n

1)

(21 + + 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 transformation 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 + + a2i(i 1) = 1 a2ii

i

n

1

:

Since C is inscribed in S we have for r = 1 a point A1 2 C \ S which musthave jA1 j = 1 since it lies in S , so a211 = 1 and the statement holds. Assume thestatement for r = m 1 < n: We observe that the following ellipsoid(8.10) (1 + )n (m

1)

(x21 + + x2m 1 ) + (1 + + 2 ) (m

p

1)

(x2m + + x2n ) 1;

where > 0; haspm 1 half axes of length (1 + )(m 1) n and n (m 1) halfaxes of length (1 + + 2 )m 1 . This implies that the volume of the ellipsoidis 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 a21 + + a2n 1:27

Thus

A() satisfies

(8.11)(1+ )n (m

[(1 + )n (m

1)

(a21 + + a2m 1 )+(1+ + 2 ) (m

1)

(a2m + + a2n ) (a21 + + a2n ) =

1](a21 + + a2m 1 ) + [(1 + + 2 ) (m

1)

1](a2m + + a2n ) 0:

1)

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 (m

1)

1] = (n

[(1+ + 2 ) m+1) 1] = (hence as

m + 1) + 2

m +1)( + 2 )+

(n

m + 1)(n m + 1) 12!

( + 2 )2 (

+

m + 1)(( m + 1) 1)2!

;

+

;

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

(8.12)

m + 1)(a21 + + a2m 1 ) + ( m + 1)(a2m + + a2n ) 0:

We now recall that we have assumed existence of A1 ; : : : ; Am 1 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 ; : : : ; Am 1 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(n

m+1)(a2m1 + +a2m(m 1) )+(m 1)a2mm = (n m+1)(1 a2mm ) (m 1)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 considerthe square of the distance of the point 1 A1 + + r Ar to the originrrXX

(8.13)

j =1

i=j

!2

i aij

rXj =1

0

2 @2j a2jj +

rXi=j +1

!2 1i aij A ;

2222 2wherePrwe have used 0 (b + c) =) b + c 2jbcj; with b = j ajj andc = i=j +1 i aij : Next we apply the CauchySchwartz theorem to see that theright hand side of (8:13) is less than or equal to(8.14)01

rXj =1

2

j ajj +2 2

rX

i=j +1

!

i

2

rX

k=j +1

!!

akj2

2

rXi=1

@a2ii +

1;kr min(iXX

k=2

j =1

1)

a2kj A 2i ;

where we have switched the main summation from j to i: In more detailPr we firstfix i = : This gives us the following terms containing the factor , k=j +1 a2kj .28

Here we must haverX

rX

j =1

i=j +1

!

i

2

j = 1; : : : 1 but also j k!rrrXXXX22akj =ik=2 j k

i=1

k=j +1

Now we use (8:9), and define the term for

a2ii +

1;kr min(iXX

k=1

j =1

1)

a2kj 1+

i

akj =2

1;kr min(iXX

1

1)

a2kj 2i :

j =1

k=2

k = 1 to be zero, to see that

rXk=1

1;j

1; hence

(a2k1 + + a2k(k

1) ) 1+

rXkk=1

n

1

= 1+

r(r

n

1)

:

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

2+r

r

n

1

Xri=1

2i := 2 :

Thus the point 1 A1 + + r Ar 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 (DvoretzkyRogers) Let B be a Banach space of infinite dimensionand let c1 ; : : : ; cr be positive numbers. Then there exist points x1 ; : : : ; xr in B suchthat kxi k2 = ci ; i = 1; : : : r and such that(8.15)where

P0

k

X0

X0xi k 2 3 c i ;

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

Proof. Choose n = r(r 1): Since B is of infinite dimension there exist n linearlyindependent elements z1 ; : : : ; zn : Then C = f(u1 ; : : : ; un ) : ku1 z1 + +un zn k 1gis a closed convex set with the origin as center in n-dimensional Euclidian space.Let A1 ; : : : ; Ar be the points provided by the DvoretzkyRogers lemma, and set

xi = c1i =2 (ai1 z1 + + ain zn ); i = 1; : : : ; r:

2Since A1 ; : : : ; Ar are on the boundary of C , we have kxi k = ci ; i = 1; : : : ; r:

P 1=2P0Furthermore the point 0 ci Ai lies in C where 2 = 2 + rr((rr 1)ci ; thus1)(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 ; : : : ; un g be a basis for E: Then for eachcan writeXn (!) = Xn1 (!)u1 + + Xnr (!)ur ;

n 2 N; !

2 we

jwhich for eachR 1 j j r gives us a real-valued asymptotic martingale, (Xn )nN ,with supn2N jXn jdP < 1: Hence strong convergence almost everywhere ofthe sequence (Xn )n2N is equivalent to coordinatewise convergence. This in turn

29

follows from the theorem on realvalued asymptotic martingales which we havealready proved.R(ii) =) (iii): Since kXn (! )k 1 for all n 2 N; ! 2 =) sup 2T kX kdP 0 we haveX dP ! 0;

2n

P

;

nN

pp 3n

( 2)

: We

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

A 2 [n2N Fn =)Furthermore we have

Z

(8.17)

A

Z

A

Xn dP ! 0:

kXn(!)k = 1; n 2 N; ! 2 ; thus

Xn dP ! 0; for all A 2 F : = ([n2N Fn ):

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

Xn (!)dP =

A

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

This together with (8:17) implies

Z

A

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

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

9

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