I have problems in understanding a small part in a proof, which is, however, a really important part.
Given:
- $X,Y,Z$ are random variables such that $\sigma(X,Y)$ and $\sigma(Z)$ are independent
- $h: \mathbb{R} \rightarrow \mathbb{R}^+$ is a Borel function which we assume to be bounded
- (*) By the uniqueness theorem of probability measures, we know that for all $A \in \mathcal{B}(\mathbb{R})$, $\mathbb{E}[h(X)\mathbb{I}_{\{(Y,Z)\in A\}}] = \mathbb{E}[\mathbb{E}[h(X)\mid Y]\mathbb{I}_{\{(Y,Z)\in A\}}] $ holds
What the proof states (not relevant for the question, only if someone is interested :-) ):
$\mathbb{P}$ a. s. $\mathbb{E}[h(X)\mid\sigma(Y,Z)] = \mathbb{E}[h(X)\mid Y]$
My question:
After (*), it is stated that by the uniqueness of the conditional expectation$^1$, $$\mathbb{E}[h(X)\mid\sigma(Y,Z)] = \mathbb{E}[h(X)\mid Y] \hspace{2cm} \tag 1$$ is implied $\mathbb{P}$ a.s.. I don't understand where the equation (1) comes from...
My attempt of explanation:
If I look at (*) and consider the uniqueness theorem of conditional expectation, I would say that $$\mathbb{E}[h(X)\mid Y] \mbox{ is a version of the conditional expectation of } h(X) \hspace{2cm} \tag 2$$ $\mathbb{P}$ a.s.. Now, I try to come from $(2)$ to $(1).$ But that does not really work. So, I think I miss something?
A further question (not so important)
Does someone may have an intuitive explanation about why $h$ maps to $\mathbb{R}^+$ and not $\mathbb{R}$? I don't see why this general case should not hold. (But this is not my main worry :-) )
Thanks a million in advance for your help!
$^1$The uniqueness theorem of conditional expectations sates that if (*) holds for two random variables $X_0$ and $\tilde X_0$, then $X_0$ = $\tilde X_0$ $\mathbb{P}$ a.s.
Definitions
A version of $E[h(X)|Y]$ is a random variable of the form $g(Y)$ (for some function $g$) that satisfies $$ E[h(X)1_{\{Y \in A\}}] = E[g(Y)1_{\{Y \in A\}}] \quad \forall A \in B(\mathbb{R})$$
A version of $E[h(X)|Y, Z]$ is a random variable of the form $r(Y,Z)$ (for some function $r$) that satisfies $$ E[h(X)1_{\{(Y,Z) \in B\}}] = E[r(Y,Z)1_{\{(Y,Z) \in B\}}] \quad \forall B \in B(\mathbb{R}^2)$$
The given problem
Now I think your problem is this: You are told $E[h(X)|Y]$ is a particular version of the conditional expectation of $h(X)$ given $Y$ (so you can call it $g(Y)$ for some function $g$ if you want, that is, $g(Y)=E[h(X)|Y]$). You are also told $E[h(X)|Y]$ satisfies an additional property called Property *:
\begin{align} &\mbox{Property *}: \\ &E[h(X)1_{\{(Y,Z) \in B\}}] = E\left[\underbrace{E[h(X)|Y]}_{g(Y)}1_{\{(Y,Z) \in B\}}\right] \quad \forall B \in B(\mathbb{R}^2) \end{align}
Finally, from this property, you are being asked to verify that your random variable $g(Y)$ is also a version of $E[h(X)|Y,Z]$.
Solution
What do we need to show to conclude that $g(Y)$ is a version of $E[h(X)|Y,Z]$? Well, we would need to show $g(Y)$ is a pure function of $(Y,Z)$ (which it is, since it is in fact a pure function of $Y$ alone) and $$ E[h(X)1_{\{(Y,Z) \in B\}}] = E[g(Y)1_{\{(Y,Z)\in B\}}] \quad \forall B \in B(\mathbb{R}^2)$$ But the given Property * itself is the same as this equation that we need to show. So this problem really just asks to verify that property * immediately implies that $g(Y)$ is a version of $E[h(X)|Y,Z]$.