Compactness and (global) convergence in measure

Question

Let $B$ denote the unit ball of $L^\infty$. Question: is $B$ sequentially compact for the topology of convergence in measure ? I am not necessarily assuming that the measure is finite (but $\sigma$ finite is fine).

(I have looked a bit, and a counterexample to a lot of similar questions was the sequence of characteristic functions $\chi_{I_n}$ where $I_n$ are intervals of length going to zero but such that $\cup_{n \geq N} I_n$ covers the real line for all $N$. However it seems to me that such a sequence does converge to $0$ in measure, so that is at least a good sign to me! )

Answer 1

With $\sigma$-finite there is a counterexample, which shows that the ball is not sequentially compact: consider $L^\infty(\mathbb R)$ with Lebesgue measure, and let $I_n=(n,n+1)$, $n\in\mathbb Z$, and $f_n=\chi^{\phantom{I_n}}_{I_n}$. Then $$ \mu(\{x:\ |f_m-f_n|\geq\varepsilon\})=2 $$ for any $\varepsilon\leq1$.

Answer 2

Even in the finite measure case this seems false.

Take $f_n(x)=\mathrm{sign}(2^n x)$ in $L^\infty(0,2\pi)$. Then for all $n \neq m$, $\mathrm{Leb}\{ x , |f_n(x)-f_m(x)|>1\}=\pi$.