Space of bounded and almost everywhere continuous function is complete?

Question

There is a well-known result that the space of bounded and continuous function is complete under sup norm.

However, I was wondering if the same result would hold if it changes for the space of bounded and almost everywhere continuous function, given for those measure 0 points are not defined.

Answer 1

Some elementary facts.

Given a set $X$ and a metric space $(Y,d)$, the space $B(X,Y)$ of bounded functions $X\to Y$ is a metric space with distance $d_\infty(f,g)=\sup_{x\in X}d(f(x),g(x))$; a sequence converges in $d_\infty$ if and only if it converges uniformly on $X$. $(B(X,Y),d_\infty)$ is complete if and only if $(Y,d)$ is. If $(Y,d)$ is an $\Bbb F$-vector space, then $d(0,\bullet)$ is a norm on $(Y,d)$ if and only if $d_\infty(0,\bullet)$ is a norm on $(B(X,Y),d_\infty)$: in this case, we usually adopt the more comfortable notation $(Y,\lVert \bullet\rVert)$ and $(B(X,Y),\lVert \bullet\rVert_\infty)$, with $\lVert f\rVert_\infty=\sup_{x\in X}\lVert f(x)\rVert$; we may also omit to specify $Y=\Bbb R$ or $Y=\Bbb C$ when it's clear from context, and just speak of $B(X)$.

All this machinery comes into play because, since we are working inside of a larger complete metric space, we only need to prove that if a sequence of almost-everywhere continuous functions converges in the $\lVert\bullet\rVert_\infty$ norm, then its limit is almost-everywhere continuous. In point of fact, we can become more ambitious and prove this:

Let $(X,\tau)$ be a topological space, $(Y,d)$ a metric space and let $f_n\to f$ in $(B(X,Y),d_\infty)$. Then, $f$ is continuous at all points where the $f_n$-s are frequently continuous as $n\to\infty$.

Proof: Consider any $x\in X$ such that for all $m\in\Bbb N$ there is some $n\ge m$ such that $f_n$ is continuous at $x$. Given any $\varepsilon>0$, there is some $n$ such that $d_\infty(f,f_h)<\varepsilon/3$ for all $h\ge n$. By hypothesis, there are some $j\ge n$ and some $U\in\tau$ such that $d(f_j(t),f_j(x))<\varepsilon/3$ for all $t\in U$. But then, for all $t\in U$ we have \begin{align}d(f(t),f(x))&\le d(f(t),f_j(t))+d(f_j(t),f_j(x))+d(f_j(x),f(x))\le\\&\le 2d_\infty(f_j,f)+d(f_j(x),f_j(t))<\varepsilon\end{align}

Now, it's fairly obvious that the result holds, because $\limsup$ of a sequence of co-null sets is co-null.