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

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.

• Hint for a counterexample: consider a "fat Cantor set" $E$ of nonzero measure. Look at a sequence of step functions (or even continuous functions) that converge pointwise to $1_E$. Show that the convergence is not locally uniform at any point of $E$. Jul 12, 2022 at 16:59

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.