Does a pointwise convergent sequence of functions converges locally uniformly almost everywhere?

Question

Does a pointwise convergent sequence of functions $f_n:\mathbb R\to \mathbb R$ converges locally uniformly almost everywhere? Here locally uniform convergence at $x$ means $x$ has a open neighborhood on which the restriction of $f_n$ converges uniformly (as opposed to the restriction of $f_n$ to every bounded set converges uniformly).

The motivation of this question is the statement of Egorov's Theorem in Terry Tao's Introduction to Measure Theory:

Let ${f_n: {\bf R}^d \rightarrow {\bf C}}$ be a sequence of measurable functions that converge pointwise almost everywhere to another function ${f: {\bf R}^d \rightarrow {\bf C}}$, and let ${\epsilon > 0}$. Then there exists a Lebesgue measurable set ${A}$ of measure at most ${\epsilon}$, such that ${f_n}$ converges locally uniformly to ${f}$ outside of ${A}$.

The author comments one cannot pick $A$ to have measure zero if one uses the definition of local uniform convergence in terms of bounded subset. But I think it is more natural to use the definition in terms of open neighborhood and I wonder if in that case the statement can be upgraded. Note that the moving bump counterexample doesn't work (since it converges locally uniformly everywhere) as well as the function $f_n=\frac{1}{nx}$ with $f_n(0)=0$ since you can just minus the point zero.

Answer 1

Not necessarily.

Consider a Smith-Volterra-Cantor set, a compact and nowhere dense set $C \subset [0,1]$ with positive Lebesgue measure. Let $C_n$ be the finite union of closed intervals remaining at stage $n$ of the construction, so that $C_1 \supset C_2 \supset \dots$ and $C = \bigcap_n C_n$.

Set $f_n = 1_{C_n}$ and $f = 1_C$ to the corresponding indicator functions, so we have $f_n \to f$ pointwise everywhere. Now I claim that for each $x \in C$, we do not have $f_n \to f$ locally uniformly at $x$. Let $U$ be any neighborhood of $x$ and let $n$ be arbitrary. Since $x \in C_n$, which is a union of closed intervals, we have $x \in I \subset C_n$ for some closed interval $I$. But $C$ is nowhere dense, so there exists some $y \in (I \setminus C) \cap U$. Hence we have $y \in U$, $f_n(y) = 1$, and $f(y) = 0$, so $\sup_{z \in U} |f_n(z) - f(z)| = 1$, and so the sequence $f_n$ does not converge to $f$ uniformly on $U$. This holds for all $x \in C$, and $C$ has positive measure.