Vrftsjtryk

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!

I think your integral is
$$ h(x) = sum_{thetain Theta} f(x,theta) pi({theta}|x) quad forall x in X $$
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 $x$.