Infinite series of integrals of $L^2$ functions I'm hoping someone can help me with this integration problem I've been struggling with.
Let $\{f_n\}$ be a sequence in $L^2(\mathbb{R})$ such that $\sum_{n=1}^\infty \lVert f_n\rVert^2_2<\infty$ and $\sum_{n=1}^\infty f_n (x)=0$ for a.e. $x\in\mathbb{R}$. I need to show that for every $g\in L^2(\mathbb{R})$,
$$\sum_{n=1}^\infty \int_\mathbb{R} f_ng\,d\mu$$
exists and is equal to zero, where $\mu$ is Lebesgue measure.
The assumptions of the problem lead me to believe that some sort of Cauchy-Schwartz Inequality result can be used, maybe along with Dominated Convergence Theorem. However, every attempt I've made at using CSI has led to unwanted square roots.
I was, however, able to prove with Fatou's Lemma that $\sum_{n=1}^\infty |f_n|^2$ is integrable. I'm not sure if it helps, but I thought I'd share it in case it does.
I'd appreciate any help for this problem.
 A: The claim is not true (we will even show that it is very false :).
To see this, let $\left(g_{n}\right)_{n\in\mathbb{N}}$
be any sequence in $L^{2}\left(\mu\right)$ which satisfies
$\sum_{n=1}^{\infty}g_{n}=0$ pointwise almost everywhere. To avoid
trivialities, also assume $\left\Vert g_{n}\right\Vert _{L^{2}}>0$
for all $n$.
Now, for each $n\in\mathbb{N}$ let
$$
k_{n}:=\left\lceil n^{2}\cdot\left\Vert g_{n}\right\Vert _{L^{2}}^{2}\right\rceil \in\mathbb{N},
$$
so that we have
$$
\sum_{\ell=1}^{k_{n}}\left\Vert \frac{g_{n}}{k_{n}}\right\Vert _{L^{2}}^{2}=\frac{1}{k_{n}^{2}}\cdot\sum_{\ell=1}^{k_{n}}\left\Vert g_{n}\right\Vert _{L^{2}}^{2}=\frac{\left\Vert g_{n}\right\Vert _{L^{2}}^{2}}{k_{n}}\leq\frac{1}{n^{2}}.
$$
Finally, define a new sequence $\left(f_{n}\right)_{n\in\mathbb{N}}$
of $L^{2}$ functions so that we have
$$
\left(f_{n}\right)_{n}=\left(\underbrace{\frac{g_{1}}{k_{1}},\dots,\frac{g_{1}}{k_{1}}}_{k_{1}\text{ terms}},\frac{g_{2}}{k_{2}},\dots,\frac{g_{2}}{k_{2}},\frac{g_{3}}{k_{3}},\dots,\frac{g_{3}}{k_{3}},\dots\right).
$$
It is then not hard to see that we have 
$$
\sum_{n=1}^{\infty}f_{n}=\sum_{n=1}^{\infty}\sum_{\ell=1}^{k_{n}}\frac{g_{n}}{k_{n}}=\sum_{n=1}^{\infty}g_{n}=0\text{ almost everywhere}.
$$
This is surely justified if the series $\sum_{n=1}^{\infty}g_{n}$
converges absolutely almost everywhere, but even if this is not the
case, it should hold. For simplicity, let us assume that the convergence
is absolute.
By our choice of $k_{n}$, we have (as seen above)
$$
\sum_{n=1}^{\infty}\left\Vert f_{n}\right\Vert _{L^{2}}^{2}=\sum_{n=1}^{\infty}\sum_{\ell=1}^{k_{n}}\left\Vert \frac{g_{n}}{k_{n}}\right\Vert _{L^{2}}^{2}\leq\sum_{n=1}^{\infty}\frac{1}{n^{2}}<\infty.
$$
If the claim was true, this would imply
$$
\int\left(\sum_{n=1}^{N}f_{n}\right)\cdot g\,{\rm d}\mu\xrightarrow[N\to\infty]{}\sum_{n=1}^{\infty}\int f_{n}\cdot g\,{\rm d}\mu
$$
for all $g\in L^{2}\left(\mu\right)$, i.e. weak convergence of $F_{N}:=\sum_{n=1}^{N}f_{n}$
to $0$ in $L^{2}\left(\mu\right)$.
But the uniform boundedness principle implies that weakly convergent
sequences are bounded (in $L^{2}$), so that we get
$$
\left\Vert \sum_{n=1}^{N}f_{n}\right\Vert _{L^{2}}\leq C
$$
for all $N\in\mathbb{N}$ and some constant $C>0$. By taking $N=k_{1}+\dots+k_{\ell}$
for arbitrary $\ell\in\mathbb{N}$, this implies in particular
$$
\left\Vert \sum_{n=1}^{\ell}g_{n}\right\Vert _{L^{2}}\leq C
$$
for all $\ell$ and thus also
$$
\left\Vert g_{\ell+1}\right\Vert _{L^{2}}=\left\Vert \left(\sum_{n=1}^{\ell+1}g_{n}\right)-\left(\sum_{n=1}^{\ell}g_{n}\right)\right\Vert _{L^{2}}\leq2C
$$
for all $\ell\in\mathbb{N}$.
But it is easy to see that we can choose the $\left(g_{n}\right)_{n\in\mathbb{N}}$
from the beginning with $\left\Vert g_{n}\right\Vert _{L^{2}}\xrightarrow[n\to\infty]{}\infty$,
for example
$$
g_{2n-1}=\chi_{\left[n,2n\right]}\text{ and }g_{2n}=-g_{2n-1}\text{ for all }n\in\mathbb{N}.
$$
Note that the convergence of the (pointwise) series $\sum_{n}g_{n}=0$
is absolute in this case, since only finitely many terms are nonzero
for each fixed $x\in\mathbb{R}$.
