Past exam of the integration course of the University of Oxford / 2017 / 2 / b / ii

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.

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.