= Integration of a series with summable L1 norms
{title2=$\int\sum_n a_n=\sum_n\int a_n$}
Suppose $\sum_n\|a_n\|_1<\infty$. By <summable L1 norms imply almost-everywhere absolute convergence>, let $A=\sum_na_n$ where the series converges absolutely, and choose zero on the exceptional <null set>. Then $A$ is a <Lebesgue measurable function> and $|A|\leq S=\sum_n|a_n|$ almost everywhere. Since $S$ is <Lebesgue integrable>, so is $A$. The signed partial sums are bounded in absolute value by $S$, so the <dominated convergence theorem> proves termwise integration. It also gives the quantitative <L1 norm> estimate
$$
\left\|A-\sum_{n=1}^Na_n\right\|_1\leq\sum_{n>N}\|a_n\|_1\longrightarrow0.
$$
The numerical series of integrals is absolutely convergent because $|\int a_n|\leq\int|a_n|$.
For example, on $(0,1)$ take $a_n(x)=(-1)^n n\mathbf1_{(0,2^{-n})}(x)$. Its <L1 norm> is $n2^{-n}$, and these norms sum to two. Every positive $x$ lies in only finitely many supports, so the series is pointwise finite, although its terms are not uniformly small. Termwise integration gives $\int_0^1\sum_na_n=\sum_n(-1)^nn2^{-n}=-2/9$, by differentiating the <geometric series>. This illustrates why a small <L1 norm> can permit tall terms on short intervals.
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