 Solution

Apply monotone convergence to $A_n={\sum }_{k=1}^n|a_k|$. Its limit $A={\sum }_{k\ge 1}|a_k|$ has integral ${\sum }_k\int |a_k|<\infty$, so $A<\infty$ almost everywhere. Thus the original series converges absolutely almost everywhere. Its partial sums are bounded in absolute value by the integrable function $A$, and their limit $s$ satisfies $|s|\le A$. Dominated convergence givesAssigning any value on the null exceptional set defines an integrable representative of $s$.