Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space. Let $Z:\Omega\to[0,\infty)$ be a bounded random variable. Let $(\mathcal{W},d)$ be a metric space and $W:\Omega \to \mathcal{W}$ be a random variable. Suppose that the differentiation theorem holds w.r.t. each measurable bounded $\varphi: \mathcal{W}\to[0,+\infty)$ and $\mathbb{P}_W$, i.e.
- for $\mathbb{P}_W$-a.e. $w \in \mathcal{W}$ it holds that $\forall r>0$, $\mathbb{P}_W(\bar{B}_r(w))>0$;
- $\frac{1}{\mathbb{P}_W \big(\bar{B}_r (w)\big)} \int_{\bar{B}_r (w)} \varphi \operatorname{d}\mathbb{P}_W \to \varphi(w), r \to 0^+$ for $\mathbb{P}_W$-a.e. $w \in \mathcal{W}$;
where, for each measurable $A \subset \mathcal{W}$, $\mathbb{P}_W$ is the probability measure defined on the Borel subsets of $(\mathcal{W},d)$, whose value at $A$ is $\mathbb{P}(W\in A)$, and $\bar{B}_r(w)$ is the closed ball of radius $r$ centered in $w$.
Is it true that
\begin{equation*}
\lim_{r\to 0^+} \frac{\mathbb{E}_{\mathbb{P}}[Z \cdot \mathbb{I}_{\bar{B}_r(w)}(W)]}{\mathbb{P}_W(\bar{B}_r(w))}
\end{equation*}
exists for $\mathbb{P}_W$-a.e. $w \in \mathcal{W}$ and that
\begin{equation*}
\Bigg(w\mapsto \lim_{r\to 0^+} \frac{\mathbb{E}_{\mathbb{P}}[Z \cdot \mathbb{I}_{\bar{B}_r(w)}(W)]}{\mathbb{P}_W(\bar{B}_r(w))}\Bigg) \circ W = \mathbb{E}_{\mathbb{P}}[Z|W]?
\end{equation*}
I proved that the results holds true if we can find a measurable space $(\mathcal{V},\mathcal{F}_{\mathcal{V}})$ and a random variable $V: \Omega \to \mathcal{V}$ such that $V$ and $W$ are $\mathbb{P}$-independent and there exists a measurable $f:\mathcal{V}\times \mathcal{W} \to [0,+\infty)$ such that $Z = f(V,W)$. However, there exist cases where this type of factorization isn't possible, so… what about the general case?
Best Answer
If you assume $Z$ is $W$-measurable, that is $Z=f(W)$ for a suitable measurable $f$, the results are direct consequences of the hypoteses.
For the more general case, that is, without assuming $Z$ to be $W$-measurable, set $Y:=\mathbb{E}_{\mathbb{P}}[Z|W]$. Then:
\begin{equation*} \mathbb{E}_{\mathbb{P}}[Z \cdot \mathbb{I}_{\bar{B}_r(w)}(W)] =\mathbb{E}_{\mathbb{P}}[Y \cdot \mathbb{I}_{\bar{B}_r(w)}(W)] \end{equation*}
for every $r>0, w \in \mathcal{W}$. This means you can replace $Z$ with $Y$ in the limit computation without altering its value (nor its existence). Remembering $Y$ is $W$-measurable by definition, you can easily conclude the proof.