Integral of a continuous function continuous on a compact set w.r.t. the pointwise convergence topology

Question

Take $K:[0,1]\times[0,1]\to [l,h] \subseteq \mathbb{R}$. For every $\tau\in [0,1]$, $K(\cdot,\tau)$ is continuous. Let $X$ be some compact subset of $[0,1]^{[0,1]}$. Does it follow that the map $$ x \mapsto \int_0^1 K(x(\tau),\tau) d\tau $$ is continuous w.r.t. the pointwise convergence topology on $X$ (assuming the integral exists for every $x\in X$)?

Answer 1

Upon further reflections, I realized ideas similar to those in my second comment lead to a counterexample assuming the continuum hypothesis. Martin's Axiom is sufficient for this purpose as well, but for simplicity I'll just work with CH. Again, I'll set $K(x, \tau) = x$. Under CH, we identify $[0, 1]$ with $\omega_1$. Let $X$ consist of all indicator functions of intervals in $\omega_1$ of the form $[0, \omega)$ for $\omega \leq \omega_1$ (or, to put it another way: $X$ consists of indicator functions of all initial segments of $\omega_1$). One can then see that $X$ is closed in $[0, 1]^{[0, 1]}$, so it is compact. Now, if $x \in X$ is the indicator function of $[0, \omega)$ for $\omega < \omega_1$, then it is supported on a countable set, so $\int_0^1 K(x(\tau), \tau) d\tau = \int_0^1 x(\tau) d\tau = 0$. The only other element of $X$ is the indicator function of the entire $[0, 1]$, so its integral is 1. But the indicator function of $[0, \omega)$ converges to the indicator function of the entire set as $\omega \rightarrow \omega_1$, showing that your map is not continuous.