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

Question

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 $$\int_E f(x)=\int_E \sum_{n=1}^\infty f_n(x)=\sum_{n=1}^\infty \int_E f_n(x).$$

Answer 1

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 $E$.