1.5. Modes of convergence 105 whenever n ≥ 1 and E is a measurable set with μ(E) ≤ δ. Show that the sequence fn is uniformly integrable. The dominated convergence theorem (Theorem 1.4.48) does not have an analogue in the uniformly integrable setting: Exercise 1.5.15. Give an example of a sequence fn of uniformly integrable functions that converge pointwise almost everywhere to zero, but do not converge almost uniformly, in measure, or in L1 norm. However, one does have an analogue of Exercise 1.5.9: Theorem 1.5.13 (Uniformly integrable convergence in measure). Let fn : X → C be a uniformly integrable sequence of functions, and let f : X → C be another function. Then fn converges in L1 norm to f if and only if fn converges to f in measure. Proof. The “only if” part follows from Exercise 1.5.2, so we establish the “if” part. By uniform integrability, there exists a finite A 0 such that X |fn| dμ ≤ A for all n. By Exercise 1.5.6, there is a subsequence of the fn that converges pointwise almost everywhere to f. Applying Fatou’s lemma (Corollary 1.4.46), we conclude that X |f| dμ ≤ A, thus f is absolutely integrable. Now let ε 0 be arbitrary. By uniform integrability, one can find δ 0 such that (1.15) |fn|≤δ |fn| dμ ≤ ε for all n. By monotone convergence, and decreasing δ if necessary, we may say the same for f, thus (1.16) |f|≤δ |f| dμ ≤ ε. Let 0 κ δ/2 be another small quantity (that can depend on A, ε, δ) that we will choose a bit later. From (1.15), (1.16) and the hypothesis κ δ/2 we have |fn−f|κ |f|≤δ/2 |fn| dμ ≤ ε

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