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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 $[0,1]$?

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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 $f_n(t) \to 1\neq f(t)$.

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