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 $d(f,g) = \int_X|f(x)-g(x)|d\mu(x)$?

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 $1/2$.

istrue 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 Lowther May 20 '12 at 22:00