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

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.

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.