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Chapter 4 Modes of Convergence

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Chapter 4 Modes of Convergence 1 CHAPTER 4 MODES OF CONVERGENCE Introduction In this chapter we will treat a variety of di¤erent sorts of convergence notions in measure theory. So called L2-convergence is of particular importance. 4.1. Convergence in Measure, in L1(); and in L2() Let (X; ; ) be a positive measure space and denote by (X) the class of measurableM functions f :(X; ) (R; ). For any f F(X); set M ! R 2 F f 1= f(x) d(x) k k j j ZX and 2 f 2= f (x)d(x): k k sZX The Cauchy-Schwarz inequality states that fg d f 2 g 2 if f; g (X): j j k k k k 2 F ZX To prove this, without loss of generality, it can be assumed that 0 < f 2< and 0 < g 2< : k k 1 k k 1 We now use the inequality 1 ( 2 + 2); ; R 2 2 2 to obtain f g 1 f 2 g2 j j j j d ( 2 + 2 )d = 1 f 2 g 2 2 f g ZX k k k k Z k k2 k k2 and the Cauchy-Schwarz inequality is immediate. If not otherwise stated, in this section p is a number equal to 1 or 2: If it is important to emphasize the underlying measure f p is written f p; : We now de…ne k k k k p () = f (X); f p< : L f 2 F k k 1g The special case p = 1 has been introduced earlier. We claim that the following so called triangle inequality holds, viz. p f + g p f p + g p if f; g (): k k k k k k 2 L The case p = 1; follows by -integration of the relation f + g f + g : j jj j j j To prove the case p = 2; we use the Cauchy-Schwarz inequality and have f + g 2 f + g 2 k k2kj j j jk2 = f 2 +2 fg d+ g 2 k k2 j j k k2 ZX 2 2 2 f +2 f 2 g 2 + g = ( f 2 + g 2) k k2 k k k k k k2 k k k k and the triangle inequality is immediate. Suppose f; g p(): The functions f and g are equal almost everywhere 2 L with respect to if f = g : This is easily seen to be an equivalence relation and the setf of6 allg equivalence 2 Z classes is denoted by Lp(): Below we consider the elements of Lp() as members of p() and two members of Lp() are identi…ed if they are equal a.e. [] : FromL this convention it is straight-forward to de…ne f + g and f for all f; g Lp() and R and (p) p2 2 the function d (f; g) = f g p is a metric on L (): Convergence in the metric space Lp() = (Lkp(),d(p)k) is called convergence in Lp(): A sequence (fk)1 in (X) converges in measure to a function f (X) if k=1 F 2 F lim ( fk f > ") = 0 all " > 0: k !1 j j 3 If the sequence (fk)k1=1 in (X) converges in measure to a function f (X) as well as to a functionFg (X); then f = g a.e. [] since 2 F 2 F " " f g > " f fk > fk g > fj j g j j 2 [ j j 2 n o n o and " " ( f g > ") ( f fk > ) + ( fk g > ) j j j j 2 j j 2 for every " > 0 and positive integer k: A sequence (fk)k1=1 in (X) is said to be Cauchy in measure if for every " > 0; F ( fk fn > ") 0 as k; n : j j ! ! 1 By the Markov inequality, a Cauchy sequence in Lp() is Cauchy in measure. p Example 4.1.1. (a) If fk = k 0; 1 ; k N+; then [ k ] 2 1 fk 2;m= 1 and fk 1;m= : k k k k pk 1 2 Thus fk 0 in L (m) as k but fk 9 0 in L (m) as k : ! ! 1 ! 1 (b) L1(m) * L2(m) since 1 2 1 [1; [(x) L (m) L (m) 1 x 2 n j j and L2(m) * L1(m) since 1 (x) L1(m) L2(m): ]0;1] x 2 n j j p Theorem 4.1.1. Suppose p = 1 or 2: (a) Convergence in Lp() implies convergence in measure: 4 (b) If (X) < ; then L2() L1() and convergence in L2() implies convergence in L1(1): p Proof. (a) Suppose the sequence (fn)n1=1 converges to f in L () and let " > 0: Then, by the Markov inequality, 1 p 1 p ( fn f ") fn f d = fn f j j "p j j "p k kp ZX and (a) follows at once. (b) The Cauchy-Schwarz inequality gives for any f (X); 2 F ( f 1d)2 f 2d 1d j j ZX ZX ZX or f 1 f 2 (X) k k k k and Part (b) is immediate. p Theorem 4.1.2. Suppose fn (X); n N+: 2 F 2 (a) If (fn)n1=1 is Cauchy in measure, there is a measurable function f : X R such that fn f in measure as n and a strictly increasing ! ! ! 1 sequence of positive integers (nj)1 such that fn f a.e. [] as j . j=1 j ! ! 1 (b) If is a …nite positive measure and fn f (X) a.e. [] as ! 2 F n ; then fn f in measure. ! 1 ! (c) (Egoro¤’s Theorem) If is a …nite positive measure and fn f (X) a.e. [] as n ; then for every " > 0 there exists E such! that2 F(E) < " and ! 1 2 M sup fk(x) f(x) 0 as n : k n j j! ! 1 x Ec 2 PROOF. (a) For each positive integer j; there is a positive integer nj such that j j ( fk fl > 2 ) < 2 ; all k; l nj: j j 5 There is no loss of generality to assume that n1 < n2 < ::: : Set j Ej = fn fn > 2 j j j+1 j and Fk = 1 Ej: [j=k If x F c and i j k 2 k fn (x) fn (x) fn (x) fn (x) j i j j j l+1 l j j l<i X l j+1 2 < 2 j l<i X c and we conclude that (fnj (x))j1=1 is a Cauchy sequence for every x Fk : Let c 2 G = 1 F and note that for every …xed positive integer k; [k=1 k 1 c j k+1 (G ) (Fk) < 2 = 2 : j=k X c Thus G is a -null set. We now de…ne f(x) = limj fnj (x) if x G and f(x) = 0 if x = G: !1 2 2 We next prove that the sequence (fn)n1=1 converges to f in measure. If x F c and j k we get 2 k j+1 f(x) fn (x) 2 : j j j Thus, if j k j+1 k+1 ( f fn > 2 ) (Fk) < 2 : j j j Since " " ( fn f > ") ( fn fn > ) + ( fn f > ) j j j j j 2 j j j 2 if " > 0; Part (a) follows at once. (b) For each " > 0; ( fn f > ") = ]"; [( fn f )d j j 1 j j ZX 6 and Part (c) follows from the Lebesgue Dominated Convergence Theorem. (c) Set for …xed k; n N+; 2 1 Ekn = 1 fj f > : [j=n j j k We have 1 Ekn Z \n=1 2 and since is a …nite measure (Ekn) 0 as n : ! ! 1 k Given " > 0 pick nk N+ such that (Ekn ) < "2 : Then, if E = 1 Ekn , 2 k [k=1 k (E) < ". Moreover, if x = E and j nk 2 1 fj(x) f(x) : j j k The theorem is proved. Corollary 4.1.1. The spaces L1() and L2() are complete. p PROOF. Suppose p = 1 or 2 and let (fn)n1=1 be a Cauchy sequence in L (): We know from the previous theorem that there exists a subsequence (fnj )j1=1 which converges pointwise to a function f (X) a.e. [] : Thus, by Fatou’s Lemma, 2 F p p f fk d lim inf fn fk d j j j j j j ZX !1 ZX p p and it follows that f fk L () and, hence f = (f fk) + fk L (): 2 2 Moreover, we have that f fk p 0 as k : This concludes the proof of the theorem. k k ! ! 1 2 2 Corollary 4.1.2. Suppose n N(0; n); n N+; and n in L (P ) as n : Then is a centred Gaussian2 random2 variable. ! ! 1 7 2 PROOF. We have that 2= E = n and 2 2=def k n k n k n k !k k as n : q Suppose! 1 f is a bounded continuous function on R. Then, by dominated convergence, E [f( )] = f(nx)d (x) f(x)d (x) n 1 ! 1 ZR ZR as n . Moreover, there exists a subsequence (nk )k1=1 which converges to a.s.! 1 Hence, by dominated convergence E f( ) E [f()] nk ! as k and it follows that ! 1 E [f()] = f(x)d 1(x): ZR By using Corollary 3.1.3 the theorem follows at once. Theorem 4.1.3. Suppose X is a standard space and a positive -…nite Borel measure on X.
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