Show that $f$ is summable on $A$ and $\lim_{n\rightarrow \infty} \int_A f_n dm=\int_A f dm$

Question

Show that if $f_n$ is summable on a bounded measurable set $A$ for $n=1,2,\ldots$ and if $f_n$ converges uniformly to $f$ on $A$ then $f$ is summable on $A$ and $\lim_{n\rightarrow \infty} \int_A f_n dm=\int_A f dm$.

Proof: It is enough to show that $\lim_{n\rightarrow \infty} \int_A (f_n-f) dm=0$. We have

$$\Bigg| \int_A(f_n-f)dm\Bigg| \le \int_A|f_n-f|dm\le m(A)\sup_{x\in A} |f_n(x)-f(x)|$$

but $\sup_{x\in A} |f_n(x)-f(x)|$ goes to $0$ as $n\rightarrow\infty$ so

$$\Bigg| \int_A(f_n-f)dm\Bigg| \rightarrow 0 \mbox{.}$$

Is this proof correct? I think I have not proved that $f$ is summable. How can I do it?

Answer 1

The proof that $\lim_{n\rightarrow \infty} \int_A f_n dm=\int_A f dm$ is correct. For the other part, take $n_0$ such that $\sup_{x\in A}|f_{n_0}(x)-f(x)|\leqslant 1$; then $$\int_A|f(x)|\mathrm dm(x)\leqslant \int_A\left(\left|f(x)-f_{n_0}(x)\right|\right)\mathrm dm+\int_A\left|f_{n_0}(x)\right|\mathrm dm(x).$$ Now I this you can finish the proof.