Continuously summing a family of maps. Let $X$ be a space and
$$\{f_i:X\rightarrow[0,\infty)\}_{i\in\mathcal{I}}$$
a family of continuous maps $X\rightarrow [0,\infty)$ indexed by some set $\mathcal{I}$. Assume that the family is point-finite. That is, that for each $x\in X$, there is a finite subset $E\subseteq\mathcal{I}$ such that $f_k(x)=0$ for $k\not\in E$. Assume furthermore that for each $x\in X$, the sum
$$\sum_{\mathcal{I}}f_i(x):=\sup_{E\subseteq\mathcal{I}}\;\sum_{i\in E}f_i(x)$$
is finite.

Under these conditions, is it true that the assignment
  $$f:x\mapsto \sum_{i\in\mathcal{I}}f_i(x)$$
  defines a continuous map $f:X\rightarrow[0,\infty)$?

I can show that $f$ is continuous if for each $x\in X$ there is an $\epsilon>0$ for which there exists a neighbourhood $U$ of $x$ and a finite $E\subseteq\mathcal{I}$ such that $\sum_{\mathcal{I}\setminus E}f_i<\epsilon$ throughout $U$. In particular it is easy to see that $f$ is continuous if $\{f_i\}_\mathcal{I}$ is locally-finite. What I can't show is that the point-finite assumption is sufficient.
 A: This can probably be made neater, but I think it works as a counterexample.
Take $X = [0, 3],$ $\mathcal{I} = \mathbb{N} = \{1, 2, 3, \ldots \},$ and
$$
f_n(x) =
\begin{cases}
nx & (0 \leqslant x \leqslant \frac1n),\\
1 & (\frac1n \leqslant x \leqslant \frac2n),\\
n\left(\frac3n - x\right) & (\frac2n \leqslant x \leqslant \frac3n), \\
0 & (\frac3n \leqslant x \leqslant 3).
\end{cases}
$$
For all $n \in \mathbb{N},$ $f_n$ is a continuous function $[0, 3] \to [0, \infty),$ and $f_n(0) = 0.$
For all $x \in (0, 3],$ $f_n(x) = 0$ for all $n$ not belonging to the finite set $\{1, 2, \ldots, \left\lfloor\frac3x\right\rfloor\}.$
Therefore the sum $f(x) = \sum_{n \in \mathbb{N}}f_n(x)$ is finite, for all $x \in [0, 3],$ and $f(0) = 0.$
But $f_n(x) \geqslant 0$ for all $n \in \mathbb{N}$ and all $x \in [0, 3],$ and
$$
f_n(x) = 1 \text{ if } n \in \left[\frac1x, \frac2x\right].
$$
A closed interval of length $a > 0$ contains at least $\left\lfloor{a}\right\rfloor$ integers, therefore
$$
f(x) \geqslant \left\lfloor\frac1x\right\rfloor \text{ for all } x \in (0, 3],
$$
whence $\lim_{x \to 0+}f(x) = +\infty,$ so the function $f \colon [0, 3] \to [0, \infty)$ is discontinuous at $0.$
