1.3. The Lebesgue integral 63 On a domain such as Rd\A, bounded-set locally uniform convergence implies open-neighbourhood locally uniform convergence, but not conversely.) Proof. By modifying fn and f on a set of measure zero (that can be ab- sorbed into A at the end of the argument) we may assume that fn converges pointwise everywhere to f, thus for every x ∈ Rd and m 0 there exists N ≥ 0 such that |fn(x) − f(x)| ≤ 1/m for all n ≥ N. We can rewrite this fact set-theoretically as ∞ N=0 EN,m = ∅ for each m, where EN,m := {x ∈ Rd : |fn(x) − f(x)| 1/m for some n ≥ N}. It is clear that the EN,m are Lebesgue measurable, and are decreasing in N. Applying downward monotone convergence (Exercise 1.2.11(ii)) we conclude that, for any radius R 0, one has lim N→∞ m(EN,m ∩ B(0,R)) = 0. (The restriction to the ball B(0,R) is necessary, because the downward monotone convergence property only works when the sets involved have finite measure.) In particular, for any m ≥ 1, we can find Nm such that m(EN,m ∩ B(0,m)) ≤ ε 2m for all N ≥ Nm. Now let A := ∞ m=1 ENm,m ∩ B(0,m). Then A is Lebesgue measurable, and by countable subadditivity, m(A) ≤ ε. By construction, we have |fn(x) − f(x)| ≤ 1/m whenever m ≥ 1, x ∈ Rd\A, |x| ≤ m, and n ≥ Nm. In particular, we see for any ball B(0,m0) with an integer radius, fn converges uniformly to f on B(0,m0)\A. Since every bounded set is contained in such a ball, the claim follows. Remark 1.3.27. Unfortunately, one cannot in general upgrade local uni- form convergence to uniform convergence in Egorov’s theorem. A basic example here is the moving bump example, fn := 1[n,n+1] on R, which “es- capes to horizontal infinity”. This sequence converges pointwise (and locally uniformly) to the zero function f ≡ 0. However, for any 0 ε 1 and any n, we have |fn(x) − f(x)| ε on a set of measure 1, namely on the interval [n, n + 1]. Thus, if one wanted fn to converge uniformly to f outside of a set A, then that set A has to contain a set of measure 1. In fact, it must contain the intervals [n, n + 1] for all suﬃciently large n and must therefore have infinite measure.

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