104 1. Measure theory (iii) (No escape to width infinity) One has sup n |fn|≤δ |fn| 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| 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 ∞. 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| 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| ε
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