I started learning about modern analysis, and I encountered the following question:

Let $(X,M,\mu)$ be a measure space, and $\|f_n-f\|_{\infty}\to 0$. Prove that there exists a set $E$ s.t $f_n-f\to 0$ uniformly and $\mu(E^c)=0$.

My attempt: convergence in $L_{\infty}$ is equivalent to: for every $n>0$ there exists $N_n$ s.t for every $m>N_n, |f_m(x)-f(x)|<\frac{1}{n}$. So, I defined $E=\bigcap_{n=1}^{\infty}\bigcup_{N_n}\bigcap_{m=N_n}^{\infty}\{x:|f_m(x)-f(x)|<\frac{1}{n}\}$. Now, if I'm not wrong uniform convergence on $E$ follows immediately, but I can't prove that $\mu(E^c)=0$. Not even sure that this is the way to go.

Any hint would be helpful.

  • $\begingroup$ Maybe it is linked with Egorov's theorem ! $\endgroup$
    – Maman
    Oct 25, 2021 at 20:35
  • 1
    $\begingroup$ @Maman Suspected that, but Egorov's theorem gives us compliment with positive measure. This is supposed to be easier. $\endgroup$
    – GBA
    Oct 25, 2021 at 20:36
  • $\begingroup$ Indeed maybe you can write explicitly $E^{c}$ ? $\endgroup$
    – Maman
    Oct 25, 2021 at 20:38
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    $\begingroup$ @Maman $E^c=\bigcup_{n=1}^{\infty}\bigcap_{N_n=1}^{\infty}\bigcup_{m=N_n}^{\infty}\{x:|f_n(x)-f(x)|\geq\frac{1}{n}\}$. Now I suspect that this is just the union of intersection of union of measure zero sets. $\endgroup$
    – GBA
    Oct 25, 2021 at 20:39
  • $\begingroup$ @JustDroppedIn No I mean convergence in $L^{\infty}$. This is regarding the first answer on this post: math.stackexchange.com/questions/187290/… $\endgroup$
    – GBA
    Oct 25, 2021 at 21:04


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