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Let $\{f_n: X\rightarrow \mathbb{R}\}$ be a sequence of continuous real-valued functions on a complete metric space, $X$. Suppose this sequence has a pointwise limit, $f$. How easy is it to see that there must be some point of $X$ at which $f$ is continuous?

I tried to use Baire Category with my open sets the sets at which the oscillation of f is bounded above by some number decreasing to zero with te index of the sets. But I couldn't show these were dense. Am I on the right track? Thanks for reading and any help would be greatly appreciated. I know that this issue has been studied and that the proof of this fact is definitely in a number of books that I don't have access to so I would be perfectly happy with an answer that just indicates how trivial or non trivial this fact is.

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Not on an arbitrary metric space, but this may help:… – Byron Schmuland Sep 28 '12 at 19:48
The points of continuity of a pointwise limit of continuous functions (= Baire class 1 function: form a comeager $G_\delta$. See Kechris, Classical Descriptive Set Theory, Theorem 24.24, page 193:… – commenter Sep 28 '12 at 20:16
Thanks for the references, guys – Leray Hirsch Sep 29 '12 at 7:53
up vote 6 down vote accepted

$\newcommand{\cl}{\operatorname{cl}}$Let $V$ be any non-empty open set in $X$.

Fix $\epsilon>0$. For $n,k\in\Bbb N$ let $G_n(k)=\{x\in V:|f_n(x)-f_{n+k}(x)|>\epsilon\}$, and let $G_n=\bigcup_{k\in\Bbb N}G_n(k)$; $G_n$ is open in $X$. Let $x\in V$; there is an $n(x)\in\Bbb N$ such that $|f_n(x)-f(x)|<\frac{\epsilon}2$ whenever $n\ge n(x)$, so $|f_{n(x)}(x)-f_{n(x)+k}(x)|<\epsilon$ for all $k\in\Bbb N$, and therefore $x\notin G_{n(x)}$. Thus, $\bigcap_{n\in\Bbb N}G_n=\varnothing$. But $V$ is a Baire space, so there must be some $n\in\Bbb N$ such that $G_n$ is not dense in $V$; let $U=V\setminus\cl_XG_n$.

Fix $x\in U$. Since $f_n$ is continuous, there is a $\delta>0$ such that $|f_n(y)-f_n(x)|<\epsilon$ for all $y\in B(x,\delta)$, the open ball in $X$ of radius $\delta$ centred at $x$, and we may further assume that $\cl_XB(x,\delta)\subseteq U$. Thus, for each $y\in B(x,\delta)$ we have


so the oscillation of $f$ on $\operatorname{cl}_XB(x,\delta)$ is at most $3\epsilon$.

Thus, for any non-empty open $V\subseteq X$ and any $\epsilon>0$ there is a non-empty open $W$ such that $\cl_XW\subseteq V$ and the oscillation of $f$ on $W$ is at most $\epsilon$. Moreover, we can make the diameter of $W$ as small as we like. It follows that for any non-empty open $V_0\subseteq X$ there is a sequence $\langle V_n:n\in\Bbb N\rangle$ of non-empty open subsets of $X$ such that for each $n\in\Bbb N$ we have $\cl_XV_{n+1}\subseteq V_n$, the diameter of $V_{n+1}$ is less than $2^{-n}$, and the oscillation of $f$ on $V_{n+1}$ is at most $2^{-n}$. $X$ is a Baire space, so $$\{x\}=\bigcap_{n\in\Bbb N}\cl_XV_n=\bigcap_{n\in\Bbb N}V_n\subseteq V_0$$ for some $x\in X$, and clearly $f$ is continuous at $x$. Thus, the set of points of continuity of $f$ is dense in $X$.

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+1 Nice answer! – user38268 Sep 29 '12 at 1:15
Thanks, Brian!! I guess when I said "variation" I meant oscillation. – Leray Hirsch Sep 29 '12 at 7:51
@Leray: You’re welcome! – Brian M. Scott Sep 29 '12 at 7:52
@BrianM.Scott nice answer! but I fail to see why after picking $n$ (so that $G_n$ is not dense) we might assume that $|f_n(x)-f(x)| < \epsilon$ for $x\in U$. From the definition, I only see that there is a $k$ such that $|f_n(x)-f_k(x)| < \epsilon$... – Yul Otani Aug 24 '13 at 17:36
@Yul: No, you get $|f_n(x)-f_{n+k}(x)|\le\epsilon$ for all $k\in\Bbb N$, and hence in the limit $|f_n(x)-f(x)|\le\epsilon$ as well. (The strictness of the inequality at the end of the displayed line comes from the middle term.) – Brian M. Scott Aug 24 '13 at 19:05

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