= Solution
Let $A_N=\sum_{n=1}^Na_n$, using the finite representatives from part (B). Then $A_N\to A$ <almost everywhere>, and $|A_N|\leq S=\sum_n|a_n|$, where $S$ is integrable by part (A). The <dominated convergence theorem> gives
$$
\boxed{\int_E\sum_{n=1}^\infty a_n(x)\,dx
=\lim_{N\to\infty}\int_EA_N
=\sum_{n=1}^\infty\int_Ea_n(x)\,dx.}
$$
The numerical series on the right converges absolutely because $|\int_Ea_n|\leq\int_E|a_n|$ and the sum of these bounds is finite. Moreover,
$$
\|A-A_N\|_1\leq\sum_{n>N}\int_E|a_n|\longrightarrow0.
$$
Thus <integration of a series with summable L1 norms> also yields convergence of the partial sums in <L1 norm>.
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