Let $D(\mathbb{R})$ be the set of all probability distributions over $\mathbb{R}$.
Is the expectation function $\mathbb{E}[\cdot]:D(\mathbb{R}) \to \mathbb{R}$ defined by $$ \mathbb{E}[\mu] := \int_{x \in \mathbb{R}} x*\mu(dx) $$ measurable? If so, how can one prove this?
To make sure my question is well-defined, I equip $\mathbb{R}$ and $D(\mathbb{R})$ with $\sigma$-algebras below. If there is an answer for different $\sigma$-algebras, that might be interesting for me as well.
- Let $\mathcal{B}$ be the Borel $\sigma$-algebra on $\mathbb{R}$
- Let the $\sigma$-algebra on $D(\mathbb{R})$ be the smallest $\sigma$-algebra containing the sets $\{ d \in D(\mathbb{R}) \mid d(B) \in B' \}$ for all $B,B' \in \mathcal{B}$.
$\mathbb{E}[\cdot]$ is only a partial function, so I have to restrict the domain of $\mathbb{E}[\cdot]$ to those probability distributions $\mu$ for which $\mathbb{E}[\mu]$ is defined. Defined in this case also means finite, but I am also fine with an answer to the related case where $\mathbb{E}[\cdot]:D(\mathbb{R}) \to \mathbb{R} \cup \{ \infty \}$.
Consider lemma Peter :
if $g:\mathbb{R}\rightarrow \mathbb{R}^+$ is measurable, then $\mu \mapsto \mathbb{E}g(\mu)=\int g d\mu$ will be $D(\mathbb{R})\rightarrow \overline{\mathbb{R}}^+$ measurable.
with the help of this lemma, you may prove your desired result using the hints/procedure written below
Write if you get stuck.
Regarding the $\sigma$-algebra: It is the standard algebra of choice, although usually it is defined by point 2 above i.e. "the $\sigma$-algebra generated by the projections". Indeed, it turns out that this is the borel algebra generated by the topology of weak convergence, so you can consider the space of probability measures as a complete and separable metric space. There is a nice appendix in D. J. Daley, D. Vere-Jones point process book about it, if you are further interested.