Summable L1 norms imply almost-everywhere absolute convergence (${\sum }_n\Vert a_n{\Vert }_1<\infty ⟹{\sum }_n|a_n(x)|<\infty \text{ a.e.}$)

Suppose measurable $a_n$ on a Lebesgue measurable set $E$ satisfy ${\sum }_n{\int }_E|a_n|<\infty$. The nonnegative partial sums $S_N={\sum }_{n\le N}|a_n|$ increase to a measurable extended-valued $S$. The monotone convergence theorem gives ${\int }_{E}S={\sum }_n{\int }_E|a_n|<\infty$, forcing $S<\infty$ almost everywhere. Each $a_n$ is finite almost everywhere too; discard the union of these countably many null sets when defining signed partial sums. The series ${\sum }_n{a}_n$ then converges absolutely outside a null set.

Suppose ${\sum }_n\Vert a_n{\Vert }_1<\infty$. By summable L1 norms imply almost-everywhere absolute convergence, let $A={\sum }_n{a}_n$ where the series converges absolutely, and choose zero on the exceptional null set. Then $A$ is a Lebesgue measurable function and $|A|\le S={\sum }_n|a_n|$ 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 estimateThe numerical series of integrals is absolutely convergent because $|\int a_n|\le \int |a_n|$.

For example, on $(0,1)$ take $a_n(x)=(-1{)}^{n}n{1}_{(0,2^{-n})}(x)$. Its L1 norm is $n{2}^{-n}$, 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 ${\int }_0^1{\sum }_n{a}_n={\sum }_n(-1{)}^{n}n{2}^{-n}=-2/9$, by differentiating the geometric series. This illustrates why a small L1 norm can permit tall terms on short intervals.