Prove that an integral of a sequence of functions converging to $0$ implies that the sequence converges to $0$ $\mu$-a.e.

Question

We are given a decreasing sequence $(f_n)_{n=1}^\infty$ of non-negative Lebesgue integrable functions on a measure space $(X,\mathcal{A},\mu)$. We are asked to show that if the integral $\int f_n \,d\mu$ converges to $0$ as $n\rightarrow\infty$ then the sequence $(f_n)_{n=1}^\infty$ converges to 0 $\mu$-a.e.

Does anyone have any ideas on how to solve this? My first thought was to use the dominated convergence theorem, since we can use $f_1$ as the integrable majorant. We also talked about the fact that the sequence converges to $0$ in mean and thus converges to $0$ in measure, which implies that there is a subsequence of $(f_n)_{n=1}^\infty$ that then converges to $0$ $\mu$-a.e. One would then need to show that there are no subsequences not converging to $0$ $\mu$-a.e.

Answer 1

Why not proceed by contradiction and note that \begin{equation*} H=\{x\in X \mid f_{n}(x) \text{ does not converge to } 0\} = \bigcup_{n=1}^{\infty}H_{m}=\bigcup_{m=1}^{\infty} \left\{x\in X \;\middle\vert\; \lim_{n\to\infty}f_{n}(x)\geq\frac{1}{m}\right\}\end{equation*} and note further that if each set in the union is $\mu$-null, then so is $H$. It follows that there is some $m_{0}$ such that $\mu(H_{m_{0}})>0$. Now, \begin{equation*} \lim_{n\to\infty}\int_{X}f_{n}(x)d\mu\geq \lim_{n\to\infty}\int_{H_{m_{0}}}f_{n}(x)d\mu\geq\int_{H_{m_{0}}}\frac{1}{m_{0}}d\mu=\frac{\mu(H_{m_{0}})}{m_{0}}>0\end{equation*} by dominated convergence.