If$f_n$ a sequence of integrable functions on a set of finite measure which converge to f uniformly, then the limit and integral can be swapped.

Question

Hello I came across this problem and tried to prove it, I would just like to confirm that my proof is correct please as I think there is something wrong with it.

The problem says the following: Suppose $(f_n)$ is a sequence of integrable functions on a set A of finite measure. Show that if the sequence is uniformly convergent on A, then $\lim_{n\to \infty} \int_{A} f_n dm= \int_{A} \lim_{n\to \infty} f_n dm$.

My attenpt: Let $\epsilon>0$ be fixed but arbitrary. Then by uniform convergence,there exists an N s.t. $|f_n-f|<\epsilon$ for all n>N. Then $0\leq \int_A |f_n-f| \leq \epsilon m(A)$ for all n>N.

Now since $\epsilon>0$ arbitrary and $m(A)<\infty$ then $\int_A |f_n-f| \to 0$ which implies $0<|\int_{A} (f_n-f)| \leq \int_{A} |f_n-f|$ and now taking the limit gives the desired result.

Could someone verify my proof? I am not sure when steps like since epsilon is arbitrary work.Also where did I use integrability of the $f_n$ ? Thanks.

Answer 1

The proof seems correct. The question doesn't make sense without the integrability of $f_n$, as the sequence on the LHS wouldn't be well-defined. Additionally, you should assert that for any $n >0$, $|f-f_n|$ is measurable, bounded, and thus, for finite measure spaces, integrable.