Assume that you are on a finite measure space $(\Omega, \mathcal{A},\mu)$. Assume that the series $\sum\limits_{i=0}^Mf_n(x)$, converges in $L^2$ to $\sum\limits_{i=0}^\infty f_n(x)$.
$\sum\limits_{i=0}^\infty f_n(x)$ is just a notation for the $L^2$ object. I do not know if the sum makes sense in a pointwise way, it is just the $L^2$ limit of the series above.
Since the measure space is finite, we also know that all our objects are in $L^1$ aswell, and that the convergence is in $L^1$ aswell.
We then can show that as a sequence of real numbers $\sum\limits_{i=0}^M\int_\Omega f_n(x)d\mu\rightarrow\int_\Omega \sum\limits_{i=0}^\infty f_n(x)d\mu$. This follows since $|\sum\limits_{i=0}^M\int_\Omega f_n(x)d\mu-\int_\Omega \sum\limits_{i=0}^\infty f_n(x)d\mu|=|\int_\Omega[\sum\limits_{i=0}^M f_n(x)d\mu-\sum\limits_{i=0}^\infty f_n(x)]d\mu|$
$\le\int_\Omega|\sum\limits_{i=0}^M f_n(x)d\mu-\sum\limits_{i=0}^\infty f_n(x)|d\mu$. And then we use the $L^1$ convergence.
So in this case, it makes sense to talk about the convergence of $\sum\limits_{i=0}^M\int_\Omega f_n(x)d\mu$ as a sequence of real numbers.
But do we know if $\sum\limits_{i=0}^M\int_\Omega f_n(x)d\mu$ converges absolutely?, that is does: $\sum\limits_{i=0}^M|\int_\Omega f_n(x)d\mu|$ converge? And does the larger sum $\sum\limits_{i=0}^M\int_\Omega |f_n(x)|d\mu$ converge? Or are there counter-examples that show that this may not be the case?