I have been looking into measure theory (from a probabilist's perspective), and I have found the proof of the existence of the conditional expectation to feel a little "glossed over" in literature. As such I've tried to very, very slowly break down the steps — and as a result I have three questions pertaining to what feels like "conceptual issues".
I will first show how I would slowly develop the proof (please correct me if I'm wrong), which will then lead to the questions at the end:
Step 1: Establish probability space: $(\Omega, \Sigma, \mathbb{P})$. Consider a $\Sigma$-measurable random variable (RV), over this space as $X$ (which would work for example as: $X: \Omega \rightarrow \mathbb{R}$)
Step 2: Define sub $\sigma$-algebra: $\mathcal{G}\subset \Sigma$, resulting in measurable space: $(\Omega, \mathcal{G})$.
Step 3: Define a measure, $\nu$, on $(\Omega, \mathcal{G})$. This measure is to behave as:
\begin{align*}
\nu(G) = \int_G X d\mathbb{P} = \int_{\Omega}X\mathbf{1}_{G} d\mathbb{P}=\mathbb{E}[X\mathbf{1}_{G}], \quad \forall G\in\mathcal{G}.
\end{align*}
Step 4: Consider now the restricted probability measure, $\mathbb{P}^{\mathcal{G}}$, restricted to measurable space, $(\Omega, \mathcal{G})$, such that $\mathbb{P}(G) = \mathbb{P}^{\mathcal{G}}(G)$, $\forall G\in\mathcal{G}$. Thus, we are now considering to work probability space: $(\Omega, \mathcal{G},\mathbb{P}^{\mathcal{G}})$.
Step 5: By construction $\nu \ll \mathbb{P}^{\mathcal{G}}$. We can thus invoke Radon-Nikodym, meaning that there is a unique (a.s.), $\mathcal{G}$-measurable function, $Z$, s.t.
\begin{align*}
\nu(G) = \int_G Z d\mathbb{P}^{\mathcal{G}} = \int_{\Omega} Z \mathbf{1}_{G} d\mathbb{P}^{\mathcal{G}} = \mathbb{E}[Z\mathbf{1}_{G}], \quad \forall G\in\mathcal{G}.
\end{align*}
Step 6: We can thus conclude the following relationships:
\begin{align*}
E[X\mathbf{1}_G] = \nu(G) = \int_{\Omega} Z\mathbf{1}_G d\mathbb{P}^{\mathcal{G}} = \mathbb{E}[Z\mathbf{1}_{G}]
\end{align*}
The significance of this is that the LHS is a $\Sigma$-measurable RV, and in the RHS it is a $\mathcal{G}$-measurable RV. Moreover, $X=Z$ (a.s.) as by Radon-Nikodym $Z$ is unique (a.s.).
Questions:
Q1: As far as I can tell, this is the proof of existence for the conditional expectation. However, for me it is not immediately obvious why this should be the conditional expectation. It seems that authors simply make a claim at the end:
Therefore, $Z=\mathbb{E}[X\mid \mathcal{G}]$…
However, I don't get why that is? Especially because $\mathbb{E}[\cdot]$ seems for me to not just be a "matter of notation", because $\mathbb{E}[X] = \int Xd\mathbb{P}$ has a very precise definition! So if we declare, $Z=\mathbb{E}[X\mid \mathcal{G}]$, it seems like the definition of $\nu(G)$ should consist of an iterated integral.
Q2: In the long list of equalities in Step 6, whereby we conclude $\mathbb{E}[X\mathbf{1}_G]=\mathbb{E}[Z\mathbf{1}_G]$, is the expectation wrt the same probability measure? i.e. is it $\mathbb{P}$ on both sides? Or $\mathbb{P}$ in the case of $\mathbb{E}[X\mathbf{1}_G]$, and $\mathbb{P}^{\mathcal{G}}$ in the case of $\mathbb{E}[Z\mathbf{1}_{G}]$ … or does it simply not matter? On this case, I have seen different authors have conflicting views, and I am not sure if it is an important issue, or simply transcription error.
Q3: What is the significance that $X$ is $\Sigma$-measurable, and $Z$ is $\mathcal{G}$-measurable? I wrote down before that "it is a significant conclusion", but I don't have an intuition why this is such an important concept.
Best Answer
Typically, these authors give the definition of conditional expectations before showing their existence. Specifically, one defines $\mathsf{E}[X\mid \mathcal{G}]$ as a (i) $\mathcal{G}$-measurable random variable $Z$ s.t. (ii) $\mathsf{E}[X1_G]=\mathsf{E}[Z1_G]$ for every $G\in\mathcal{G}$. The R-N Theorem implies that such a random variable exists. In fact, it is the R-N derivative of $\nu$ w.r.t. $\mathsf{P}$.
It doesn't matter because $Z1_G$ is $\mathcal{G}$-measurable.
The significance is that in general we cannot take $Z=X$ (it satisfies (ii) trivially, but $X$ is not necessarily $\mathcal{G}$-measurable).