Past exam of the options II course of the University of Oxford / 2010 / D / D3

The monotone convergence theorem states that for nonnegative measurable functions $f_n\uparrow f$ almost everywhere, $\int f={\lim }_n\int f_n$, allowing the value $+\infty$.

The dominated convergence theorem states that if measurable functions $f_n\to f$ almost everywhere and $|f_n|\le g$ almost everywhere for some integrable $g$, then $f$ is integrable and $\int f_n\to \int f$. In fact $\int |f_n-f|\to 0$ as well. Both statements apply on any measure space.

Apply the monotone convergence theorem to $s_N={\sum }_{k=1}^N|a_k|$. Its limit $s$ satisfiesThus $s$ is finite almost everywhere, which proves absolute convergence of ${\sum }_k{a}_k$ almost everywhere. Define its sum $a$ there, and set $a=0$ on the exceptional null set. Then $|a|\le s$, so $a$ is integrable. The partial sums of ${\sum }_k{a}_k$ are also dominated by $s$; dominated convergence givesThe series of integrals converges absolutely, since $|\int a_k|\le \int |a_k|$.

For $0<x<1$, expand $(1-x{)}^{-2}={\sum }_{k\ge 1}k{x}^{k-1}$. Multiplying by $(\log x{)}^2$ gives nonnegative terms, so the monotone convergence theorem permits termwise integration. Put $x=e^{-t}$ to obtainwhere scaling followed by two integrations by parts evaluates the last integral. HenceThe finite value also establishes integrability at both endpoints.