64 1. Measure theory However, if all the fn and f were supported on a fixed set E of finite measure (e.g. on a ball B(0,R)), then the above “escape to horizontal infinity” cannot occur, it is easy to see from the above argument that one can recover uniform convergence (and not just locally uniform convergence) outside of a set of arbitrarily small measure. We now use Theorem 1.3.20 to give another version of Littlewood’s sec- ond principle, known as Lusin’s theorem: Theorem 1.3.28 (Lusin’s theorem). Let f : Rd → C be absolutely inte- grable, and let ε 0. Then there exists a Lebesgue measurable set E ⊂ Rd of measure at most ε such that the restriction of f to the complementary set Rd\E is continuous on that set. A word of caution: This theorem does not imply that the unrestricted function f is continuous on Rd\E. For instance, the absolutely integrable function 1Q : R → C is nowhere continuous, so is certainly not continuous on R\E for any E of finite measure but on the other hand, if one deletes the measure zero set E := Q from the reals, then the restriction of f to R\E is identically zero and thus continuous. Proof. By Theorem 1.3.20, for any n ≥ 1 one can find a continuous, com- pactly supported function fn such that f − fn L1(Rd) ≤ ε/4n (say). By Markov’s inequality (Lemma 1.3.15), that implies that |f(x) − fn(x)| ≤ 1/2n−1 for all x outside of a Lebesgue measurable set En of measure at most ε/2n+1. Letting E := ∞ n=1 En, we conclude that E is Lebesgue measurable with measure at most ε/2, and fn converges uniformly to f outside of E. But the uniform limit of continuous functions is continuous, and the same is true for local uniform limits (because continuity is itself a local property). We conclude that the restriction f to Rd\E is continuous, as required. Exercise 1.3.23. Show that the hypothesis that f is absolutely integrable in Lusin’s theorem can be relaxed to being locally absolutely integrable (i.e. absolutely integrable on every bounded set), and then relaxed further to that of being measurable (but still finite everywhere or almost everywhere). (To achieve the latter goal, one can replace f locally with a horizontal truncation f1|f|≤n alternatively, one can replace f with a bounded variant, such as f (1+|f|2)1/2 .) Exercise 1.3.24. Show that a function f : Rd → C is measurable if and only if it is the pointwise almost everywhere limit of continuous functions fn : Rd → C. (Hint: If f : Rd → C is measurable and n ≥ 1, show that

Purchased from American Mathematical Society for the exclusive use of nofirst nolast (email unknown) Copyright 2011 American Mathematical Society. Duplication prohibited. Please report unauthorized use to cust-serv@ams.org. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.