Let $(X,\Sigma,\mu)$ be a finite measure space (i.e., $\mu(X) < \infty$). Let $\mathcal{F}$ be the set of $\mu$-measurable functions $f:X \to \mathbb{R}$ that are bounded in $[0,1]$, so that $0 \leq f(x) \leq 1$ for all $x \in X$ and $f \in \mathcal{F}$. Is the set $\mathcal{F}$ compact with respect to the topology induced by the $L_1$ metric $d(f,g) = \int_X|f(x)-g(x)|d\mu(x)$?
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4$\begingroup$ Probably not research level, but I decided to answer anyway. Jet-lag may serve as an excuse for that --- also for any mistakes in my answer. $\endgroup$– Andreas BlassCommented May 20, 2012 at 20:39
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$\begingroup$ @Vladimir: a Banach space is locally compact iff it is finite dimensional, and $L^1(X,\Sigma,\mu)$ is finite dimensional iff $X$ is the union of finitely many atoms. $\endgroup$– Pietro MajerCommented May 20, 2012 at 21:12
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2$\begingroup$ It is true that for any uniformly bounded sequence $f_n$, there exists a sequence $g_n\in{\rm conv}(f_n,f_{n+1},\ldots)$ which converges in $L^1$. (conv=set of convex combinations). This is often a good enough replacement for compactness. It has been used many times in the papers of Delbaen and Schachermayer on no arbitrage conditions in stochastic calculus (and by the same authors to prove the Bichteler-Dellacherie theorem and the Doob-Meyer decomposition). $\endgroup$– George LowtherCommented May 20, 2012 at 22:00
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Well, if $X$ is a finite set, then yes. But in the cases you probably had in mind, no. Suppose, for example, that $X$ is $[0,1]$ with Lebesgue measure, and let $f_n(x)$ be the $n$-th digit of the binary expansion of $x$. No subsequence converges, since the $L_1$ distance between any two distinct $f_n$'s is $1/2$.
answered May 20, 2012 at 20:38