# 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!

• Thanks @Michael, you’re right. Actually, the space $\Theta$ I considered was a compact metric space, but now I just want to simplify the question and restrict it to be finite. – Paradiesvogel Nov 18 '18 at 10:50
• Thanks @Michael . I totally agree with you. In fact, I really need the probability distribution $\pi(\cdot | x)$ depending on $x$. This means for each $x \in X$, I have a different probability distribution defined on $2^\Theta$. Also, the space $\Theta$ is at least not trivial. In such a setting, what can I do to ensure the map $h(x)$ is continuous on $X$? Is it possible to do that? Thanks a million again :-) – Paradiesvogel Nov 18 '18 at 11:02

## 1 Answer

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 $$x$$.

• Thanks so much for your answer @Michael. I should clarify initially that given any fixed $x \in X$, we have a different probability distribution; that can be reviewed as for each state variable $x$ in a State space $X$, I will have a different probability distribution on $\Theta$ and such a probability distribution depends on different observations of $x$. In this setting, could we still get some result for continuity of the map $h$? Would you mind rethinking about the question based on this setting please? I sincerely appreciate your kind help! – Paradiesvogel Nov 18 '18 at 11:16
• I think this is what my example already does. Note that in this example I refined my answer to have $\Theta$ now a 2-element set rather than a 1-element set (for a 1-element set then the mass function must be 1 (in order to be a mass function) so I could not really define $\pi(\{\theta\}|x)$ discontinuously in $x$ when $\Theta$ is a 1-element set). – Michael Nov 18 '18 at 11:20
• Thanks @Michael . Did you mean that a sufficient condition for the continuity of the function $h$ is $\pi(\{\theta\} | x)$ is continuous in $x$ for each $\theta \in \Theta$? Besides, may I ask what do you think if we extend the finite space of $\Theta$ to a compact metric space? Is it possible to do that? – Paradiesvogel Nov 18 '18 at 11:31
• Yes, my first paragraph says it is sufficient to have $\pi(\{\theta\}|x)$ continuous in $x$ for each $\theta \in \Theta$, for the case when $\Theta$ is a finite set. – Michael Nov 18 '18 at 11:32
• Thank you very much @Michael . I understand now. But still curious about what if the space $\Theta$ is allowed to be compact. In this setting, since the space $\Theta$ could be infinite or countably infinite, such a sufficient condition may fail. Do you think is there any reasonable assumption imposed on $\pi$ that would guarantee the continuity of $h$ with integral? Many thanks :-) – Paradiesvogel Nov 18 '18 at 11:41