1.5. Modes of convergence 103 1.5.5. Domination and uniform integrability. Now we turn to the re- verse question, of whether almost uniform convergence, pointwise almost ev- erywhere convergence, or convergence in measure can imply L1 convergence. The escape to vertical and width infinity examples shows that, without any further hypotheses, the answer to this question is no. However, one can do better if one places some domination hypotheses on the fn that shut down both of these escape routes. We say that a sequence fn : X → C is dominated if there exists an absolutely integrable function g : X → C such that |fn(x)| ≤ g(x) for all n and almost every x. For instance, if X has finite measure and the fn are uniformly bounded, then they are dominated. Observe that the sequences in the vertical and width escape to infinity examples are not dominated (why?). The dominated convergence theorem (Theorem 1.4.48) then asserts that if fn converges to f pointwise almost everywhere, then it necessarily con- verges to f in L1 norm (and hence also in measure). Here is a variant: Exercise 1.5.9. Suppose that fn : X → C are a dominated sequence of measurable functions, and let f : X → C be another measurable function. Show that fn converges in L1 norm to f if and only if fn converges in measure to f. (Hint: One way to establish the “if” direction is to first show that every subsequence of the fn has a further subsequence that converges in L1 to f, using Exercise 1.5.6 and the dominated convergence theorem (Theorem 1.4.48). Alternatively, use monotone convergence to find a set E of finite measure such that X\E g dμ, and hence X\E fn dμ and X\E f dμ, are small.) There is a more general notion than domination, known as uniform integrability, which serves as a substitute for domination in many (but not all) contexts. Definition 1.5.11 (Uniform integrability). A sequence fn : X → C of abso- lutely integrable functions is said to be uniformly integrable if the following three statements hold: (i) (Uniform bound on L1 norm) One has sup n fn L1(μ) = sup n X |fn| dμ +∞. (ii) (No escape to vertical infinity) One has sup n |fn|≥M |fn| dμ → 0 as M → +∞.

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