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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.

Let $S_N(x)={\sum }_{n=1}^N|f_n(x)|$. These are nonnegative measurable functions increasing to the extended-valued function $S(x)={\sum }_{n=1}^{\infty }|f_n(x)|$. The monotone convergence theorem givesTherefore $S$ is finite almost everywhere: if $S=\infty$ on a measurable set of positive Lebesgue measure, its integral would exceed every finite bound. Hence the series ${\sum }_n{f}_n(x)$ converges absolutely outside the null set $N=\{S=\infty \}$. This is summable L1 norms imply almost-everywhere absolute convergence.

Define $s(x)={\sum }_n{f}_n(x)$ off $N$ and set $s=0$ on $N$. The function is Lebesgue measurable, being the pointwise limit of measurable partial sums off a measurable null set. Since $|s|\le S$ almost everywhere,so the sum is Lebesgue integrable.

For $s_N={\sum }_{n=1}^N{f}_n$, the absolute tail satisfiesApply the monotone convergence theorem to the nonnegative tail partial sums. We obtainThus the partial sums converge in L1 norm. The triangle inequality for an integral yieldsConsequentlyThe scalar series on the right also converges absolutely, since ${\sum }_n|{\int }_E{f}_n|\le {\sum }_n\Vert f_n{\Vert }_1<\infty$. This proves integration of a series with summable L1 norms without needing pointwise convergence on the exceptional null set.

Let $S_N={\sum }_{n=1}^N|a_n|$. These are nonnegative measurable functions with $S_N\uparrow S={\sum }_{n\ge 1}|a_n|$. The monotone convergence theorem and linearity of the finite sum giveHence $S<\infty$ almost everywhere: if $S=\infty$ on a set of positive measure, the integral would exceed every finite constant. Also each $a_n$ is finite almost everywhere, since its absolute integral is finite. Outside the union of these countably many null sets, all terms are real andThus the signed series converges absolutely almost everywhere, as asserted by summable L1 norms imply almost-everywhere absolute convergence. Removing the common null set avoids any undefined $+\infty -\infty$ in signed partial sums.