Prove that the normed space $L^{\infty}$ equipped with $\lVert\cdot\rVert_{\infty}$ is complete. 
Possible Duplicate:
Understanding proof of completeness of $L^{\infty}$ 

Most of the materials I have in Real Analysis consider this statement as a trivial one: "The normed space $L^{\infty}$ equipped with $\lVert\cdot\rVert_{\infty}$ is  complete".
But to my suprise I can't see the triviality. I am searching for it now...
Anybody with hint?
 A: Let $\left(f_n\right)_{n\geqslant 1}$ be a Cauchy sequence in $L^{\infty}$ endowed with the natural norm. For $n,m$, let $N_{n,m}$ a set of measure $0$ such that $|f_n(x)-f_m(x)|\leq \lVert f_n-f_m\rVert_{\infty}$ for each $x\notin N_{n,m}$. Define $N:=\bigcup_{n,m}N_{n,m}$. Then $N$ is of measure $0$ as a countable union of such sets and the sequence of functions $\left(\widetilde f_n\right)_{n\geqslant 1}$ restricted to $N^c$ is uniformly convergent. Then you can find an uniform limit $f$ (i.e. such that $\lVert f-\widetilde f_n\rVert_{\infty}\to 0$. Then just define $f$ by $0$ on $N$ to get a limit in $L^{\infty}$ (more precisely the limit will be the equivalence class of this function).
A: Edit: I have (hopefully) provided a non-formal way of vizualising what is going on here. Well $l^\infty = (C[a,b], ||.||_\infty)$ is incomplete because there are Cauchy sequences in $l^\infty$ which do not converge in $l^\infty$ (can you think of any?). So we want to "fill up" the space $l^\infty$ with functions so that all the Cauchy sequences converge. This "filled up" space is called $L^\infty$. The new vector space has all the functions in the space $l^\infty$, and also has discontinuous "jump" functions, but it still has the $\sup$ norm $\Vert\cdot\Vert_\infty$. We have "completed" the space $l^\infty$ and ended up with $L^\infty$ (and so we see that the space $L^\infty$ is complete). This is basically the definition of what $L^\infty$ is.
