Show that if $\sum_{n=1}^\infty |f_n(x)|\leq \mathcal{C}$ for almost all $x\in E$, then $f(x)=\sum_{n=1}^\infty f_n(x)$ converges

Let $E$ be a finite measurable subset of the real line and $f_n$ a sequence of integrable functions on $E$. Show that if $$\sum_{n=1}^\infty |f_n(x)|\leq \mathcal{C}$$ for almost all $x\in E$, then $$f(x)=\sum_{n=1}^\infty f_n(x)$$ converges almost everywhere on $E$.

The second part comes naturally from the first time that needs to be proven but how do you do the first?

1 Answer

Denote $f_n^\pm$ to be the positive or negative part of the function $f_n$ in the sense that $$f_n = f_n^+ - f_n^-$$ now note that since $f_n^\pm \leq |f_n|$ so for almost all $x \in E$ we have $$\sum_{n=1}^\infty f_n^\pm \leq \sum_{n=1}^\infty |f_n| \leq C$$ since $f_n^\pm$ are non-negative the partial sums form an increasing function which is bounded above and hence both the sequences are convergent for almost all $x \in E$.

Finally note that for almost all $x \in E$ $$\sum_{n=1}^\infty f_n^+ - \sum_{n=1}^\infty f_n^- = \sum_{n=1}^\infty f_n = f$$