Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 3 / 22F / c

First suppose $1\le p<\infty$. Choose a subsequence so thatBy Tonelli theorem,Hence the sum inside the integral is finite almost everywhere, which forces $|f_{n_k}(x)-f(x)|\to 0$ almost everywhere.

If $p=\infty$, choose representatives and remove the countable union of the null sets on whichOutside that null set, norm convergence bounds the pointwise difference and the entire sequence converges to $f$. Thus in every case an almost-everywhere convergent subsequence exists, as stated in Almost-everywhere convergent subsequence from Lp convergence.

Regard the set-theoretic inclusionas a linear map between Banach spaces. Its graph is closed. Indeed, if $f_n\to f$ in $L^q$ and $f_n\to g$ in $L^p$, part (i) gives an almost-everywhere convergent subsequence for the first convergence and then a further subsequence converging almost everywhere for the second; hence $f=g$ almost everywhere. The closed graph theorem therefore gives

Let $E_R=X\cap B(0,R)$. It has finite measure, and $\mu (E_R)\uparrow \mu (X)$. Applying the bound to $1_{E_R}$ gives, for finite $q$,so $\mu (E_R{)}^{1/p-1/q}\le C$. For $q=\infty$, it gives $\mu (E_R{)}^{1/p}\le C$. In either case the measures $\mu (E_R)$ are uniformly bounded. Continuity from below now yieldsThis proves the converse summarized by Lq inclusion implies finite measure on a Euclidean Borel subset.