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2. Convergence Theorems

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2. Convergence Theorems 2. Convergence theorems In this section we analyze the dynamics of integrabilty in the case when se- quences of measurable functions are considered. Roughly speaking, a “convergence theorem” states that integrability is preserved under taking limits. In other words, ∞ if one has a sequence (fn)n=1 of integrable functions, and if f is some kind of a limit of the fn’s, then we would like to conclude that f itself is integrable, as well R R as the equality f = limn→∞ fn. Such results are often employed in two instances: A. When we want to prove that some function f is integrable. In this case ∞ we would look for a sequence (fn)n=1 of integrable approximants for f. B. When we want to construct and integrable function. In this case, we will produce first the approximants, and then we will examine the existence of the limit. The first convergence result, which is somehow primite, but very useful, is the following. Lemma 2.1. Let (X, A, µ) be a finite measure space, let a ∈ (0, ∞) and let fn : X → [0, a], n ≥ 1, be a sequence of measurable functions satisfying (a) f1 ≥ f2 ≥ · · · ≥ 0; (b) limn→∞ fn(x) = 0, ∀ x ∈ X. Then one has the equality Z (1) lim fn dµ = 0. n→∞ X Proof. Let us define, for each ε > 0, and each integer n ≥ 1, the set ε An = {x ∈ X : fn(x) ≥ ε}. ε Obviously, we have An ∈ A, ∀ ε > 0, n ≥ 1. One key fact we are going to use is the following. Claim 1: For every ε > 0, one has the equality ε lim µ(An) = 0. n→∞ Fix ε > 0. Let us first observe that, using (a), we have the inclusions ε ε (2) A1 ⊃ A2 ⊃ ... T∞ ε Second, using (b), we clearly have the equality k=1 Ak = ∅. Since µ is finite, using the Continuity Property (Lemma III.4.1), we have ∞ ε \ ε lim µ(An) = µ An = µ( ) = 0. n→∞ ∅ n=1 Claim 2: For every ε > 0 and every integer n ≥ 1, one has the inequality Z ε 0 ≤ fn dµ ≤ aµ(An) + εµ(X). X 300 §2. Convergence theorems 301 Fix ε and n, and let us consider the elementary function ε h = a ε + ε ε , n κ An κ Bn ε ε ε where Bn = X r A . Obviously, since µ(X) < ∞, the function hn is elementary ε integrable. By construction, we clearly have 0 ≤ fn ≤ hn, so using the properties of integration, we get Z Z ε ε ε ε 0 ≤ fn dµ ≤ hn dµ = aµ(An) + εµ(B ) ≤ aµ(A ) + εµ(X). X X Using Claims 1 and 2, it follows immediately that Z Z 0 ≤ lim inf fn dµ ≤ lim sup fn dµ ≤ εµ(X). n→∞ X n→∞ X Since the last inequality holds for arbitrary ε > 0, the desired equality (1) immedi- ately follows. We now turn our attention to a weaker notion of limit, for sequences of mea- surable functions. Definition. Let (X, A, µ) be a measure space, let K be a one of the symbols ¯ R, R, or C. Suppose fn : X → K, n ≥ 1, are measurable functions. Given ∞ a measurable function f : X → K, we say that the sequence (fn)n=1 converges µ-almost everywhere to f, if there exists some set N ∈ A, with µ(N) = 0, such that lim fn(x) = f(x), ∀ x ∈ X N. n→∞ r In this case we write f = µ-a.e.- lim fn. n→∞ Remark 2.1. This notion of convergence has, among other things, a certain uniqueness feature. One way to describe this is to say that the limit of a µ-a.e. con- vergent sequence is µ-almost unique, in the sense that if f anf g are measurable func- tions which satisfy the equalities f = µ-a.e.- limn→∞ fn and g = µ-a.e.- limn→∞ fn, then f = g, µ-a.e. This is quite obvious, because there exist sets M, N ∈ A, with µ(M) = µ(N) = 0, such that lim fn(x) = f(x), ∀ x ∈ X M, n→∞ r lim fn(x) = g(x), ∀ x ∈ X N, n→∞ r then it is obvious that µ(M ∪ N) = 0, and f(x) = g(x), ∀ x ∈ X r [M ∪ N]. Comment. The above definition makes sense if K is an arbitrary metric space. Any of the spaces R¯, R, and C is in fact a complete metric space. There are instances where the requirement that f is measurable is if fact redundant. This is somehow clarified by the the next two exercises. Exercise 1*. Let (X, A, µ) be a measure space, let K be a complete separable metric space, and let fn : X → K, n ≥ 1, be measurable functions. (i) Prove that the set ∞ L = x ∈ X : fn(x) n=1 ⊂ K is convergent belongs to A. 302 CHAPTER IV: INTEGRATION THEORY (ii) If we fix some point α ∈ K, and we define ` : X → K by ( lim fn(x) if x ∈ L `(x) = n→∞ α if x ∈ X r L then ` is measurable. In particular, if µ(X r L) = 0, then ` = µ-a.e.- limn→∞ fn. Hints: If d denotes the metric on K, then prove first that, for every ε > 0 and every m, n ≥ 1, the set ε Dmn = x ∈ X : d fm(x), fn(x) < ε belongs to A (use the results from III.3). Based on this fact, prove that, for every p, k ≥ 1, the set p 1 Ek = x ∈ X ; d fm(x), fn(x) < p , ∀ m, n ≥ k belongs to A. Finally, use completeness to prove that ∞ ∞ \ [ p L = Ek . p=1 k=1 Exercise 2. Use the setting from Exercise 1. Prove that Let (X, A, µ), K, and ∞ (fn)n=1 be as in Exercise 1. Assume f : X → K is an arbitrary function, for which there exists some set N ∈ A with µ(N) = 0, and lim fn(x) = f(x), ∀ x ∈ X N. n→∞ r Prove that, when µ is a complete measure on A (see III.5), the function f is auto- matically measurable. Hint: Use the results from Exercise 1. We have X r N ⊂ L, and f(x) = `(x), ∀ x ∈ X r N. Prove that, for a Borel set B ⊂ K, one has the equality f −1(B) = `−1(B)4M, for some M ⊂ N. By completeness, we have M ∈ A, so f −1(B) ∈ A. The following fundamental result is a generalization of Lemma 2.1. Theorem 2.1 (Lebesgue Monotone Convergence Theorem). Let (X, A, µ) be ∞ 1 a measure space, and let (fn)n=1 ⊂ L+(X, A, µ) be a sequence with: • fn ≤ fn+1, µ-a.e., ∀ n ≥ 1; R • sup X fn dµ : n ≥ 1 < ∞. Assume f : X → [0, ∞] is a measurable function, with f = µ-a.e.- limn→∞ fn. Then 1 R R f ∈ L+(X, A, µ), and X f dµ = limn→∞ X fn dµ. R Proof. Define αn = X fn dµ, n ≥ 1. First of all, we clearly have 0 ≤ α1 ≤ α2 ≤ ..., ∞ so the sequence (αn)n=1 has a limit α = limn→∞ αn, and we have in fact the equality α = sup αn : n ≥ 1 < ∞. 1 With these notations, all we need to prove is the fact that f ∈ L+(X, A, µ), and that we have Z (3) f dµ = α. X Fix a set M ∈ A, with µ(M) = 0, and such that limn→∞ fn(x) = f(x), ∀ x ∈ X r M. For each n, we define the set Mn = {x ∈ X : fn(x) > fn+1(x)}. §2. Convergence theorems 303 Obviously Mn ∈ A, and by assumption, we have µ(Mn) = 0, ∀ n ≥ 1. Define the S∞ set N = M ∪ n=1 Mn . It is clear that µ(N) = 0, and • 0 ≤ f1(x) ≤ f2(x) ≤ · · · ≤ f(x), ∀ x ∈ X r N; • f(x) = limn→∞ fn(x), ∀ x ∈ X r N. So if we put A = X r N, and if we define the measurable functions gn = fnκ A, n ≥ 1, and g = fκ A, then we have (a) 0 ≤ g1 ≤ g2 ≤ · · · ≤ g (everywhere!); (b) limn→∞ gn(x) = g(x), ∀ x ∈ X; (c) gn = fn, µ-a.e., ∀ n ≥ 1; (d) g = f, µ-a.e.; 1 R Notice that property (c) gives gn ∈ L+(X, A, µ) and X gn dµ = αn, ∀ n ≥ 1. By property (d), we see that we have the equivalence 1 1 f ∈ L+(X, A, µ) ⇐⇒ f ∈ L+(X, A, µ). 1 R R Moreover, if g ∈ L+(X, A, µ), then we will have X g dµ = X f dµ. These observa- tions show that it suffices to prove the theorem with g’s in place of the f’s. The advantage is now the fact that we have the slightly stronger properties (a) and (b) above. The first step in the proof is the following. −1 α Claim 1: For every t ∈ (0, ∞), one has the inequality µ g ((t, ∞]) ≤ t . −1 Denote the set g ((t, ∞]) simply by At. For each n ≥ 1, we also define the set n −1 At = gn ((t, ∞]). Using property (a) above, it is clear that we have the inclusions t 2 (4) A1 ⊂ At ⊂ · · · ⊂ At. S∞ n Using property (b) above, we also have the equality At = n=1 At . Using the continuity Lemma 4.1, we then have n µ(At) = lim µ(At ), n→∞ so in order to prove the Claim, it suffices to prove the inequalities α (5) µ(An) ≤ n , ∀ n ≥ 1. t t But the above inequality is pretty obvious, since we clearly have 0 ≤ t n ≤ gn, κ At which gives Z Z n tµ(A ) = t n dµ ≤ gn dµ = αn.
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