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

Set $C_0=[0,1]$. Inductively, replace each component interval of $C_n$ by its two closed end intervals, each of length ${\ell }_{n+1}:=q_{n+1}/2^{n+1}$. The current component length is ${\ell }_n=q_n/2^n$. Because $q_{n+1}<q_n$, the removed central gap has positive lengthThis is a generalized Cantor set construction. The three required properties are checked in parts i, ii and iii.

The sets are closed, measurable, decreasing, and contained in $[0,1]$, which has finite Lebesgue measure. By continuity from above of a measure,This limit exists because $(q_n)$ is a bounded decreasing monotone sequence. To exclude a nondegenerate interval $J\subseteq C$, note that its connectedness forces it into a single component interval of every $C_n$: crossing two components would cross a removed gap. Its length would therefore be at most $2^{-n}$ for every $n$, which is impossible for a positive-length interval. Thus $C$ contains no nondegenerate interval, even when its Lebesgue measure is positive.

At stage zero there is one closed interval. At every later stage each component is replaced by two closed end intervals separated by a positive open gap. Induction therefore gives exactly $2^n$ pairwise disjoint closed intervals at stage $n$. Their common length issince the decreasing sequence starts at $q_0=1$.

The $2^n$ component intervals are pairwise disjoint and each has length $q_n/2^n$. Finite additivity of Lebesgue measure givesTheir endpoints do not cause overlaps, because the central gaps have strictly positive length.

Each interval retained at stage $n+1$ lies inside the stage-$n$ interval from which it was cut. Taking the union over all parent components proves $\boxed{C_{n+1}\subseteq C_n}$. Consequently the intersection defining $C$ is a nested intersection of nonempty compact sets.

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 $A$.

The a.e. limit $f$ is nonnegative. For a fixed measurable set $A$, Fatou's lemma gives ${\int }_{A}f\le {\mathrm{liminf}}_n{\int }_A{f}_n$. Apply the same lemma to $A^c$ and use convergence of the total integrals:The lower and upper bounds prove $\boxed{{\int }_A{f}_n\to {\int }_{A}f}$.

For the asserted L1 convergence, choose an integer $k$ so large that ${\int }_{|x|>k}f<\epsilon$, possible by Lebesgue integrability. The first result applied to this fixed tail set makes ${\int }_{|x|>k}{f}_n<2\epsilon$ eventually. The additional local hypothesis makes ${\int }_{-k}^k|f_n-f|<\epsilon$ eventually. ThereforeThus $\boxed{\Vert f_n-f{\Vert }_{L^1(\mathbb{R})}\to 0}$.

In fact the extra local hypothesis is unnecessary here. Scheffé's lemma follows directly from $\min \{f_n,f\}\to f$ almost everywhere and domination by $f$:This stronger argument uses precisely the already assumed nonnegativity, a.e. convergence, and convergence of total integrals.