My instructor briefly discussed a result in lecture that I need help with. Here was the set up. We

My instructor briefly discussed a result in lecture that I need help with. Here was the set up.
We assumed that $\left\{{f}_{n}\right\}$ is an orthonormal system in ${L}^{2}\left(0,1\right)$ and we supposed that there exists an $M>0$ such that $|{f}_{n}\left(x\right)|\le M$ a.e for all $n\in \mathbb{N}$. Furthermore, we let $\left\{{c}_{n}\right\}$ be a sequence of real numbers such that $\sum _{n=1}^{\mathrm{\infty }}{c}_{n}{f}_{n}$cnfn converges a.e.
They said that it was "clear" that
$\underset{n\to \mathrm{\infty }}{lim}{c}_{n}=0.$
This is not clear or intuitive for me.
After reading through the page, I am still a bit confused. Can someone put together a working proof to clear things up?
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Shawn Castaneda
Suppose the sum of an infinite series is $\phantom{\rule{thickmathspace}{0ex}}S\phantom{\rule{thickmathspace}{0ex}}$ ,and we denote by $\phantom{\rule{thickmathspace}{0ex}}{S}_{n}\phantom{\rule{thickmathspace}{0ex}}$ the $\phantom{\rule{thickmathspace}{0ex}}n\phantom{\rule{thinmathspace}{0ex}}-$−th element of the sequence of partial sums of the series. By definition, the series converges iff the limit of $\phantom{\rule{thickmathspace}{0ex}}{S}_{n}\phantom{\rule{thickmathspace}{0ex}}$ exists finitely ( and in this case $\phantom{\rule{thickmathspace}{0ex}}{S}_{n}\to S\phantom{\rule{thickmathspace}{0ex}}$) , so:
$S=\sum _{n=1}^{\mathrm{\infty }}{a}_{n}\phantom{\rule{thickmathspace}{0ex}},\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}{S}_{n}:=\sum _{k=1}^{n}{a}_{k}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{a}_{n}={S}_{n}-{S}_{n-1}\underset{n\to \mathrm{\infty }}{\overset{}{\to }}\stackrel{\text{(arith. of limits)}}{S}-S=0$
and we have the basic necessary condition for convergence of infinite series.
In your case you have $\phantom{\rule{thickmathspace}{0ex}}|{c}_{n}{f}_{n}|\underset{n\to \mathrm{\infty }}{\overset{}{\to }}0\phantom{\rule{thickmathspace}{0ex}}$ (no matter whether real of complex numbers), but $\phantom{\rule{thickmathspace}{0ex}}\left\{{f}_{n}\right\}\phantom{\rule{thickmathspace}{0ex}}$ is a bounded sequence...! Finish the argument.