Let $(X,M,\mu)$ be a measure space. let $f_n$ be a Cauchy sequence in $L^\infty(X)$. I want to show that there is a mesurable $E\subset X$ such that $\mu(E^c)=0$ and $\forall \epsilon \gt 0$, $|f_n-f_m| \le \epsilon$ on $E$ for some $n,m\gt N$.
Furthermore, I want to use the above to show that $L^\infty(X)$ is complete.
So I know for $n,m\gt N$ $|f_n-f_m|\le \|f_n-f_m\|\le \epsilon $ a.e. Let $E_k = \{x : |f_k(x) \gt \|f_k\|_\infty \}\qquad E_{n,m} = \{ x : |f_n (x) -f_m(x)|\gt \|f_n-f_m\|_\infty \}$ Set $E^c = \bigcup_k E_k \cup \bigcup_{n,m} E_{k,m}$ then $\mu(E^c)=0$ and so $|f_n-f_m|\leq \epsilon$ on $E$.
For the second part, since $\mathbb R$ is complete, on $E$, $f_n \to f$ uniformly.
Also, $|f(x)|\leq |f_n(x)-f(x)|+|f_n| \lt 1+ |f_n|$, so $f\in L^\infty$. I can also say that $|f_n-f|=\lim_{n\to\infty} |f_m-f_n|\le \epsilon$ on $E$.
How do I get that $\|f-f_n\|_\infty\to 0$? Is my approach right?