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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 $A(x)={\sum }_{n\ge 1}{a}_n(x)$ on the full-measure set where part (A) gives absolute convergence, and set $A=0$ on its null complement. After similarly replacing the individual terms by zero on that common null set, the partial sums are finite measurable functions and converge to $A$. Hence $A$ is a Lebesgue measurable function. The triangle inequality givesalmost everywhere. Part (A) proved $S$ integrable, soTherefore the summed function is Lebesgue integrable. Its values on the exceptional null set do not affect the integral. This is the integrability step in integration of a series with summable L1 norms.

Let $A_N={\sum }_{n=1}^N{a}_n$, using the finite representatives from part (B). Then $A_N\to A$ almost everywhere, and $|A_N|\le S={\sum }_n|a_n|$, where $S$ is integrable by part (A). The dominated convergence theorem givesThe numerical series on the right converges absolutely because $|{\int }_E{a}_n|\le {\int }_E|a_n|$ and the sum of these bounds is finite. Moreover,Thus integration of a series with summable L1 norms also yields convergence of the partial sums in L1 norm.

Use the power series of the exponential function:Each term is a Lebesgue integrable function, andThe sum of these nonnegative integrals is at most $2{\sum }_{n\ge 0}1/n!=2e<\infty$. Applying integration of a series with summable L1 norms, just proved in part (ii), therefore justifies exchanging the series and the integral:The absolute-integral bound is the justification; pointwise convergence of the exponential series alone would not suffice.