Proving an equivalence to convergence in $L_\infty$ norm

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.

• Indeed maybe you can write explicitly $E^{c}$ ? Oct 25, 2021 at 20:38
• @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.
• @JustDroppedIn No I mean convergence in $L^{\infty}$. This is regarding the first answer on this post: math.stackexchange.com/questions/187290/…