# If $f_n$ converges uniformly to $f$ on a measure space, show integral of $f_n$ converges to integral of $f$.

Let $(\Omega,\cal F, \mu)$ be a measure space on which $(f_n)$ is a sequence of bounded, measurable, real-valued functions converging uniformly to $f$.

1. If the measure of $W$ is finite, the integral of $f_n$ on $\Omega$ converges to the integral of $f$ on $W$. (Should I use monotone convergence THM or dominance convergence THM or neither?)

2. Show by an example that if the finite-measure hypothesis is dropped then the conclusion may fail.

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1. No need to use "big" theorems: $$\int_\Omega|f_n-f|d\mu\leqslant \mu(\Omega)\sup_{x\in \Omega}|f_n(x)-f(x)|.$$
2. Consider the set of positive natural numbers and $f_n=n^{-1}(\underbrace{1,\dots,1}_{n\mbox{ times}},0,\dots)$, $f\equiv 0$.
$f_n$ is a sequence in this case, whose $n$ first term are $n^{-1}$, and all the others $0$. There is uniform convergence to the null sequence, but not in $L^1$. –  Davide Giraudo Apr 13 '13 at 16:59