Counter example for limit of integrals, with continuity Let $(\Xi,\Sigma,\mu)$ be a probability space, $(X,d)$ a metric space and $f : X\times \Xi \to \mathbb{R}$ be a function such that


*

*$x \mapsto f(x,\xi)$ is continuous for all $x,\xi$

*$\xi \mapsto f(x,\xi)$ is integrable for all $x$

*$\sup_{x}\int_{\Xi}|f(x,\xi)|d\nu(\xi) < \infty$.


Is it possible that with these assumptions the function
$x \mapsto \int_{\Xi}f(x,\xi)d\mu(\xi)$ is not continuous?
My thoughts so far:
The continuity amounts to showing that
$\int_{\Xi}f(x_n,\xi)d\mu(\xi) \to \int_{\Xi}f(x,\xi)d\mu(\xi)$ for any sequence $x_n \to x$.
In general the identity $\lim_{n\to\infty}\int_{\Xi}f_n(\xi)d\mu = \int_{\Xi}\lim_{n\to\infty}f_n(\xi)d\mu(\xi)$ may not hold, but the reason I have hope for this is that the counter examples I've seen don't seem like they can be written as limits involving continuous functions, for instance the classic example of $f_n(\xi) = n\chi_{(0,1/n)}(\xi)$ on $[0,1]$
 A: Consider this simpler problem: Suppose $f_n(\xi) \to f(\xi)$ pointwise on $[0,1],$ and we have $\int_0^1|f_n| \le M$ for all $n.$ Is it true that $\int_0^1 f_n \to \int_0^1 f?$ No, here's a standard counterexample: $f_n(\xi) = n^2\xi^n(1-\xi).$ Here $f_n(\xi) \to 0$ pointwise everywhere on $[0,1],$ while $\int_0^1 f_n = n^2[1/(n+1) - 1/(n+2)] \to 1.$
I used that idea to construct a counterexample for your question on $[0,1]\times [0,1]:$
$$f(x,\xi) = \begin{cases} \frac{1}{x^2}\xi^{1/x}(1-\xi)\, d\xi,& (x,\xi) \in (0,1]\times [0,1] \\ 
0, & (x,\xi) \in \{0\}\times [0,1] \end{cases}.$$
A: Choose a continuous function $\chi$ such that its support is contained in $\left[\tfrac{1}{4},\tfrac{3}{4}\right]$ and $\int \chi(\xi) \, d\xi = 1$ (e.g. a hat function/triangular function). Consider $X=[0,1]$ (endowed with the Euclidean scalar product) and $\Xi = (0,1)$ (with the Lebesgue measure restricted to $(0,1)$). For $x \in X$ and $\xi \in \Xi$ define
$$f(x,\xi) := \begin{cases} \frac{1}{x} \chi \left( \frac{\xi}{x} \right), & x \in (0,1], \\ 0, & x=0. \end{cases}$$
Since $\chi$ is continuous and $\chi(\xi)=0$ for all $\xi \geq \tfrac{3}{4}$, the function $X \ni x \mapsto f(x,\xi)$ is continuous for any $\xi \in \Xi$. Moreover,
$$F(x) := \int_{\Xi} f(x,\xi) \, d\xi = \frac{1}{x} \int_{(0,1)} \chi \left( \frac{\xi}{x} \right) \, d\xi = \int_{(0,1/x)} \chi(\eta) \, d\eta = 1$$
for all $x \in (0,1]$. On the other hand, we have
$$F(0) = \int 0 \, d\xi = 0.$$
This means that $F$ is not continuous at $x=0$. 
