A question about continuity of a specific function with probability measure Let $X$ be a compact metric space, and $\Theta$ be a finite space, endowed with their own $\sigma$-algebra.
Let $f \colon X \times \Theta \to \mathbb{R}$ be a Caratheodory function such that
(1) for each $x \in X$, the function $f(x, \cdot) \colon \Theta \to \mathbb{R}$
is measurable; and (2) for each $\theta \in \Theta$, the function $f( \cdot, \theta) \colon X \to \mathbb{R}$ is continuous.
Given each $x \in X$, we have a probability distribution $\pi( \cdot \,| \, x) \colon 2^{\Theta} \to [0,1]$. In particular, given any fixed $x \in X$,
it will generate a corresponding probability distribution $\pi$ on $2^\Theta$.
I am curious that 

Under what kind of conditions (assumptions) imposed on this probability distribution $\pi$ , the map $$X \ni x \mapsto \int_\Theta f(x,\theta) \, \pi( \mathrm{d} \theta \,| \,x) \in \mathbb{R}$$
  will be continuous on $X$?

Any idea or suggestions are most welcome!
Thank you so much!
 A: I think your integral is 
$$ h(x) = \sum_{\theta\in \Theta} f(x,\theta) \pi(\{\theta\}|x) \quad \forall x \in X $$
if $\pi(\{\theta\}|x) = \pi(\{\theta\})$ for all $x \in X$ then this is a sum of a finite number of functions that are continuous in $x$, and hence is continuous in $x$. More generally, if $\pi(\{\theta\}|x)$ is continuous in $x$ for each $\theta \in \Theta$, then this is a sum of a finite numer of functions that are continuous in $x$ (and hence is continuous in $x$).
Else, it is easy to get a discontinuous example (despite my incorrect comment from before that tried to do it with $\Theta$ being only a 1-element set) by defining $\pi(\{\theta\}|x)$ discontinuously.  Define $X=[0,1]$, define $\Theta=\{0,1\}$, $f(x,0)=0$, $f(x,1) = 1$ for all $x \in [0,1]$, and define: 
$$ (\pi(\{0\}|x), \pi(\{1\}|x)) = \left\{ \begin{array}{ll}
(1,0) &\mbox{ if $x \in [0,1/2)$} \\
(1/2,1/2)  & \mbox{ if $x \in [1/2,1]$} 
\end{array}
\right.$$
Then 
$$h(x)= \pi(\{1\}|x) = \left\{ \begin{array}{ll}
0 &\mbox{ if $x \in [0,1/2)$} \\
1/2  & \mbox{ if $x \in [1/2,1]$} 
\end{array}
\right.$$ 
and this is discontinuous in $x$.
