102 1. Measure theory Corollary 1.5.10. Suppose that fn : X C are a sequence of measurable functions that converge in L1 norm to a limit f. Then there exists a sub- sequence fnj that converges almost uniformly (and hence, pointwise almost everywhere) to f (while remaining convergent in L1 norm, of course). Proof. Since fn −f L1(μ) 0 as n ∞, we can select n1 n2 n3 . . . such that fnj −f L1(μ) 2−j (say). This is enough for the previous exercise to apply. Actually, one can strengthen this corollary a bit by relaxing L1 conver- gence to convergence in measure: Exercise 1.5.6. Suppose that fn : X C are a sequence of measur- able functions that converge in measure to a limit f. Then there ex- ists a subsequence fnj that converges almost uniformly (and hence, point- wise almost everywhere) to f. (Hint: Choose the nj so that the sets {x X : |fnj (x) f(x)| 1/j} have a suitably small measure.) It is instructive to see how this subsequence is extracted in the case of the typewriter sequence. In general, one can view the operation of passing to a subsequence as being able to eliminate “typewriter” situations in which the tail support is much larger than the width. Exercise 1.5.7. Let (X, B,μ) be a measure space, let fn : X C be a sequence of measurable functions converging pointwise almost everywhere as n to a measurable limit f : X C, and for each n, let fn,m : X C be a sequence of measurable functions converging pointwise almost everywhere as m (keeping n fixed) to fn. (i) If μ(X) is finite, show that there exists a sequence m1,m2,... such that fn,mn converges pointwise almost everywhere to f. (ii) Show the same claim is true if, instead of assuming that μ(X) is finite, we merely assume that X is σ-finite, i.e., it is the countable union of sets of finite measure. (The claim can fail if X is not σ-finite. A counterexample is if X = NN with counting measure, fn and f are identically zero for all n N, and fn,m is the indicator function of the space of all sequences (ai)i∈N NN with an m.) Exercise 1.5.8. Let fn : X C be a sequence of measurable functions, and let f : X C be another measurable function. Show that the following are equivalent: (i) fn converges in measure to f. (ii) Every subsequence fnj of the fn has a further subsequence fnji that converges almost uniformly to f.
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