104 1. Measure theory (iii) (No escape to width infinity) One has sup n |fn|≤δ |fn| dμ → 0 as δ → 0. Remark 1.5.12. It is instructive to understand uniform integrability in the step function case fn = An1En . The uniform bound on the L1 norm then asserts that Anμ(En) stays bounded. The lack of escape to vertical infinity means that along any subsequence for which An → ∞, Anμ(En) must go to zero. Similarly, the lack of escape to width infinity means that along any subsequence for which An → 0, Anμ(En) must go to zero. Exercise 1.5.10. (i) Show that if f is an absolutely integrable func- tion, then the constant sequence fn = f is uniformly integrable. (Hint: Use the monotone convergence theorem.) (ii) Show that every dominated sequence of measurable functions is uniformly integrable. (iii) Give an example of a sequence that is uniformly integrable but not dominated. In the case of a finite measure space, there is no escape to width infinity, and the criterion for uniform integrability simplifies to just that of excluding vertical infinity: Exercise 1.5.11. Suppose that X has finite measure, and let fn : X → C be a sequence of measurable functions. Show that fn is uniformly integrable if and only if supn |fn|≥M |fn| dμ → 0 as M → +∞. Exercise 1.5.12 (Uniform Lp bound on finite measure implies uniform integrability). Suppose that X has finite measure, let 1 p ∞, and suppose that fn : X → C is a sequence of measurable functions such that supn X |fn|p dμ ∞. Show that the sequence fn is uniformly integrable. Exercise 1.5.13. Let fn : X → C be a uniformly integrable sequence of functions. Show that for every ε 0 there exists a δ 0 such that E |fn| dμ ≤ ε whenever n ≥ 1 and E is a measurable set with μ(E) ≤ δ. Exercise 1.5.14. This exercise is a partial converse to Exercise 1.5.13. Let X be a probability space, and let fn : X → C be a sequence of absolutely integrable functions with supn fn L1 ∞. Suppose that for every ε 0 there exists a δ 0 such that E |fn| dμ ≤ ε

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