Suppose that $\{f_{n}\}$ is a sequence of measurable real valued functions on $[0,1]$ and that $M>0$ is a constant such that for all $n, 0\leq f_{n}(x)\leq M$ a.e. and $\int_{0}^{1}f_{n}=1$ for all $n$. If $\{a_{n}\}$ is a sequence of positive numbers such that $\sum_{n=1}^{\infty}a_{n}f_{n}(x)<\infty $ a.e., then $\sum_{n=1}^{\infty}a_{n}<\infty.$
Let $f(x)=\sum_{n=1}^{\infty}a_{n}f_{n}(x)$. By monotone convergence theorem, $\sum_{n=1}^{\infty}a_{n}=\int_{0}^{1}f$. By continuity of measure, for each $\epsilon>0,\exists T>0$ s.t. $m(\{f(x)>T\})<\epsilon.$ For each $n$, let $T_{n}>0$ s.t. $m(\{f(x)>T_{n}\})<\frac{1}{2^{n}}$, then $$\int_{0}^{1}f=\sum_{n=1}^{\infty}\int_{\{T_{n}<f\leq T_{n+1}\}}f+\int_{\{f\leq T_{1}\}}f\leq \sum_{n=1}^{\infty}\frac{T_{n+1}}{2^{n}}+T_{1}.$$ Here I lost control of $T_{n}$. How can I continue from here?