Past exam of the integration course of the University of Oxford / 2021 / 3 / b

The points belonging to infinitely many $E_n$ form their limit superior of sets:For every $N$, countable subadditivity givesThus $\boxed{m(G)=0}$. Outside this null set, each point belongs to only finitely many $E_n$, so its indicator functions are eventually zero. Therefore $\boxed{{\chi }_{E_n}(x)\to 0\text{ almost everywhere}}$. This is the summable-measure argument behind the first Borel-Cantelli lemma; no independence assumption is needed.

By part i, ${\chi }_{E_n}|g|\to 0$ almost everywhere, while $0\le {\chi }_{E_n}|g|\le |g|$ and $g$ is Lebesgue integrable. The dominated convergence theorem givesTo deduce absolute continuity of the Lebesgue integral, suppose the stated uniform conclusion failed for some ${\epsilon }_0>0$. For each $n$ one could choose a measurable set $A_n$ with $m(A_n)<2^{-n}$ but ${\int }_{A_n}|g|\ge {\epsilon }_0$. Applying the result just proved to this sequence contradicts the lower bound. Hencefor every measurable set $A$.