Question. Let $f: \mathbb{R} \to \mathbb{R}$ be any function. For $n = 1, 2, \dots$ let $$g_n(x) = \sum_{k \in \mathbb{Z}} f(k/n) \mathbb{1}_A,$$
where $\mathbb{1}_A$ is the indicator function for $A = \left[\frac{k}{n}, \frac{k+1}{n}\right)$.
Show $g_n(c) \to f(c)$ as $n \to \infty$ if there exists some point $c \in \mathbb{R}$ where $f$ is continuous.
Let $x \in \mathbb{R}$. We assume $f$ is continuous as some point $c$, and so for all $\epsilon > 0$ there exists a $\delta > 0$ such that
$$|x - c| < \delta ~\Rightarrow~ |f(x) - f(c)| < \epsilon.$$
Now we want to show that $\lim_{n \to \infty} g_n(c) \to f(c)$.
My idea is to do something like: $$ \begin{align} \lim_{n \to \infty} g_n(c) &= \lim_{n \to \infty} \sum_{k \in \mathbb{Z}} f(k/n)\mathbb{1}_A \\ &= \sum_{k \in \mathbb{Z}} \lim_{n \to \infty} f(k/n) \mathbb{1}_A \tag{Continuous at $c$}\\ &= \sum_{k \in \mathbb{Z}} f(0) \mathbb{1}_{[0, 0)} \\ &\leq \sum_{k \in \mathbb{Z}} \frac{\epsilon}{2^k} \\ &= \epsilon. \end{align} $$
Im not sure on if my steps from, $\lim_{n \to \infty} \sum_{k \in \mathbb{Z}} f(k/n)\mathbb{1}_A = \dots \leq \sum_{k \in \mathbb{Z}} \frac{\epsilon}{2^k}$, are sufficient? It seems to me you would need to make the extra assumption that $f(c) = f(0)$, so $c = 0$ is this particular point of continuity?
