How to prove this fact about a sequence of Lebesgue integrals?

Question

I'd like some help to prove the following statement:

Let (X, M, u) be a measure space (X a set, M a sigma-algebra on X and u a measure on M) and let (f_n) be a sequence of integrable complex valued functions defined on X. If f_n converges uniformly to a function f and u(X) < oo, then f is integrable and lim Int f_n du = Int f du.
Show that the requirement that u(X) be finite is essential, that is, show that, if u(X) = oo, then it's possible that the functions f_n are integrable and converge uniformly to f and, yet, the equality lim Int f_n du = Int f du does not hold (in virtue of Fatou's Lemma, in this case we must have Int f_n du < Int f du).
Thank you for any help

Answer 1

For each n, put e_n = supremum |f_n(x) - f(x)|, the sup. being taken over X. Since f_n -> f uniformily, e_n -> 0 and e_n is eventually finite. For every n we have |f - f_n| < e_n => |f| < |f_n| + e_n. For sufficiently large n, e_n is finite. Since |f_n| is integrable - for f_n is - and u(X) < oo, it follows that Int (|f_n| + e_n) du = Int |f_n| du + e_n u(X)
From |f - f_n| <= e_n, it also follows that f_n - e_n <= f <= f + e_n . Integrating these functions, the inequalities are preserved and we get Int f_n du - e_n u(X) <= Int f du <= Int f_n du + e_n u(X) => | Int f_n du - Int f du| <= e_n u(X), valid for every n. Since u(X) is finite and e_n -> 0, e_n u(X) -> 0 and it follows , by confront, that lim | Int f_n du - Int f du| = 0. Since f is integrable, this implies lim Int f_n du = Int f du .

To show u(X) < oo is an essential condition, take X = [0, oo), M = the Lebesgue sigma- algebra and u = Lebesgue measure. For each n, put f_n =C(x)/n, where C is the characteristic function of [0, n]. It's easy to show f_n converges uniformly to f = 0, the null function. So, int f du = 0 But for every n, Int f_n du = u([0, n])/n = n/n =1 , so that lim Int f_n du =1 > 0 = Int f du . It's easy to show no functio g that dominates f_n can be integrable, because we have Int g du >= 1 + 1/2....+1/n..... = oo
Hope this helps

Answer 2

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