# If $\sum_{n=1}^\infty |f_n(x)|\leq \mathcal{C}$ for almost all $x\in E$, then $\int_E f=\sum_{n=1}^\infty \int_E f_n$.

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

We know that $$f(x)=\sum_{n=1}^\infty f_n(x)$$ converges almost everywhere on $E$. So $$\int_E f(x)=\int_E \sum_{n=1}^\infty f_n(x)=\sum_{n=1}^\infty \int_E f_n(x).$$

• There are some problems here. I think you mean that $E$ has finite measure, not that $E$ is finite. Also, you should define $f = \sum_{n = 1}^{\infty} f_n$. Finally, the statement is clearly wrong. – user296602 Apr 11 '16 at 0:57
• This is a one-line application of dominated convergence. – user296602 Apr 11 '16 at 1:00
• Simply saying "converges almost everywhere" is not enough. Use Bongers' hint. – GEdgar Apr 11 '16 at 19:55

As pointed out by T. Bongers, we can use the dominated convergence theorem defining $g_N(x):=\sum_{n=1}^Nf_n(x)$. Its absolute value is dominated by a constant independent of $N$, which is integrable over the finite measure set $E$.