$L^p$ space is complete metric (Rudin RCA) I'm reading theorem 3.11 in Rudin's RCA what says $L^p$ space is a complete metric space. 
At the end of the proof, Rudin says that "Then $\mu(E)=0$, and on the complement of $E$ the sequence ${f_n}$ converges uniformly to a bounded function $f$". 
Why ${f_n}$ converges uniformly?
I see $$E^c=\{f_n : |f_n (x)|\leq \|f_n\|_\infty \text{ and }  |f_n-f_m|\leq \|f_n-f_m\|_\infty \}$$

 A: Since $\|f_n(x)-f_m(x)| \leq \|f_n-f_m\|_{\infty}$ for all $n,m$, for $x$ not in $E$ and  since $\|f_n-f_m\|_{\infty} \to 0$ the convergence is uniform. Remember that this proof is for the case $p=\infty$ where we are starting with a Cauchy sequence in $L^{\infty}$.
A: By definition of $\|\cdot\|_\infty$, the sets
$$A_k=\{x \in X : |f_k(x)| > \|f_k\|_\infty\},\quad B_{m,n}=\{x \in X : |f_n(x)-f_m(x)| > \|f_n-f_m\|_\infty\}$$
are $\mu$-null sets so $E$ is a $\mu$-null sets as a countable union of such sets.
For every $x \in E^c$ we have
$$|f_n(x) - f_m(x)| \le \|f_n - f_m\|_\infty \xrightarrow{m,n\to\infty} 0 \quad \text{ uniformly in } x \in E^c$$
so $(f_n|_{E^c})_n$ is uniformly Cauchy. Also since $|f_n(x)| \le \|f_n\|_{\infty}, \forall x \in E^c$ we have that $f_n|_{E^c}$ are bounded functions.
Now recall that the space $B(E^c, \mathbb{R})$ of bounded functions $E^c \to \mathbb{R}$ equipped with the metric $d(f,g) = \sup_{x \in E^c}|f(x)-g(x)|$ is a complete metric space. Since $(f_n|_{E^c})_n$ is a Cauchy sequence in $B(E^c, \mathbb{R})$, there exists $f \in B(E^c, \mathbb{R})$ such that $f_n|_{E^c} \to f$ uniformly on $E^c$.
In particular we have $f_n|_{E^c} \to f$ pointwise so $f$ is a measurable function as a limit of measurable functions.
