# Convergence of a sequence of right continuous functions on a dense set

Let $$f_n : [0,1] \to \mathbb{R}$$ be a sequence of right-continuous functions and $$f : [0,1] \to \mathbb{R}$$ a right-continuous function. Assume that there exists a dense set $$D \subseteq [0,1]$$ such that $$f_n(x) \to f(x)$$ for all $$x \in D$$. Does it follow that $$f_n(x) \to f(x)$$ for all points $$x$$ where $$f$$ is continuous?

Let $$x_0 \in [0,1]$$ be a point where $$f$$ is continuous. An idea for a proof would be to split $$|f_n(x_0) - f(x_0)| \leq |f_n(x_0) - f_n(x)| + |f_n(x) - f(x)| + |f(x) - f(x_0)|$$ for some $$x \in D$$. One can choose $$x$$ such that both $$|f(x) - f(x_0)| < \frac{\varepsilon}{3}$$ (because $$f$$ is continuous in $$x_0$$) and $$|f_n(x) - f(x)| < \frac{\varepsilon}{3}$$ for $$n \geq n_0$$ (because $$f_n(x) \to f(x)$$ for all $$x \in D$$). But why is it possible to choose $$x$$ such that the third inequality $$|f_n(x_0) - f_n(x)| < \frac{\varepsilon}{3}$$ for all $$n \geq n_0$$ also holds? The right-continuity was not used so far. Do we need it at all? If right-continuity is not enough, then what if all the functions have also left-hand limits everywhere in $$[0,1]$$?

This result is false. Let $$t \in (0,1)$$. Let $$f_n$$ be piece-wise linear function such that $$f_n(x)=0$$ for $$|x-t| >\frac 1 n$$ and $$f_n(t)=1$$. Let $$f\equiv 0$$. Then $$f_n(x) \to f(x)$$ for all $$x\neq t$$ (hence on a dense set of points) but $$f_n(t) \to 1\neq f(t)$$.