 Solution

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.