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 $d(f,g) = \int_Xf(x)g(x)d\mu(x)$?
Take the 2minute tour
×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $1/2$. 

