I am trying to show that $L^\infty$ is complete, and I am struggling to show that the limiting function I have come up with is in $L^\infty$. Here is what I have so far:
We let $f_n$ be a Cauchy sequence in $L^\infty$, that is, for any $\epsilon > 0$, there exists an $N$ such that we have $$||f_n -f_m||_\infty < \epsilon,$$ for all $n,m \ge N$. Now, we fix $x$, consider $f_n(x)$, and observe that there exists some $N'$ such that $|f_n(x) - f_m(x)| < \epsilon$, for all $n,m > N'$, for if there was not, for any $k$, including $N$, we could find $n',m' > k$ such that $|f_n'(x) - f_m'(x)| \ge \epsilon$, and thus, $||f_n-f_m||_\infty \ge \epsilon$, and so the sequence $f_n$ would not be Cauchy in the $L^\infty$ norm.
Next, note that since $f_n(x)$ is Cauchy for each $x$, and $f_n(x) \in \Re$, then $f_n(x) \to y_x$, for some $y \in \Re$. Let $f(x) = y_x$. We claim that $f(x) \in L^\infty$, and that $||f_n - f||_\infty \to 0$.
To prove the first assertion, we note that Cauchy sequences in $\Re$ are bounded, this gives us that $|f_n(x)| \le M_x$ for some $M_x$.
This is where I am stuck. I don't understand how to go from here to getting that $f \in L^\infty$. I am not sure if I am going about this whole proof wrong, or making this much more complicated than it should be, but any help would be much appreciated!
Take $n$ such that $j,k \geq n$ implies $$ \|f_j - f_k\|_\infty < 1\,. $$ Now $$ \|f_j\|_\infty \leq \|f_j - f_n\|_\infty + \|f_n\|_\infty \leq 1 + \|f_n\|_\infty\,. $$ We know that $|f_j(x)| \leq \|f_j\|_\infty$ a.e. $x$ and $|f(x)| = \lim_{j \to \infty} |f_j(x)|$. What can you say about this limit now?
To show that $\|f_n - f\|_\infty \to 0$, use the fact that $$ |f_j(x)-f_k(x)| \leq \|f_j - f_k\|_\infty $$ almost everywhere, and thus $$ |f(x) - f_j(x)| = \lim_{k \to \infty} |f_j(x)-f_{j+k}(x)| \leq \lim_{k \to \infty} \|f_j - f_{j+k}\|_\infty $$ almost everywhere.
Now what do you know about $\|f-f_j\|_\infty$ once $j$ is big enough?