$\sum_{n=1}^{\infty} | f_n(x)|<\infty$ a.e. $\Rightarrow$ $\lim_{N\to \infty}\sum_{n=1}^N f_n (x)$ exist (at least a.e.)

Question

Let $\{ f_n \}_{n=1}^{\infty}$ be a sequence of functions.

If $\displaystyle\sum_{n=1}^{\infty} \big| f_n(x) \big|<\infty$ almost everywhere $x$, then does $\displaystyle\lim_{N\to \infty}\sum_{n=1}^N f_n (x)$ exist (at least a.e. $x$) ?

The fact that $\displaystyle\sum_{n=1}^{\infty} \big| f_n(x) \big|<\infty$ almost everywhere means $\displaystyle\sum_{n=1}^{\infty} f_n(x)$ abosulety convergents a.e. $x$.

I wonder whether this fact implies $\displaystyle\lim_{N\to \infty}\sum_{n=1}^N f_n (x)$ exists (at least a.e. $x$) or not.

Thank you for your help.

Answer 1

Of course, the limit exists a.e. . Fix some $x$ such that $$ \sum_{n = 1}^\infty \lvert f_n(x) \rvert < \infty. $$ Then the sequence $\displaystyle N \mapsto \sum_{n = 1}^N \lvert f_n(x) \rvert$ is increasing (trivial) and bounded (by assumption). It therefore converges (if $f$ is Banach-space-, e.g. $\mathbb{R}$-valued). You even have $f_n(x) \rightarrow 0$ a.e. .

Answer 2

If $\sum_{n=1}^{\infty}|f_n(x)| <\infty,$ then $\sum_{n=1}^{\infty}f_n(x)$ converges absolutely, hence converges. Since the first sum converges for a.e. $x,$ the second sum converges for a.e. $x.$