Basic properties of the conditional expectation in Banach spaces Let


*

*$E_1$ be a normed $\mathbb R$-vector space

*$E_2$ be a separable $\mathbb R$-Banach space

*$(\Omega,\mathcal A,\operatorname P)$ be a probability space

*$\mathcal F\subseteq\mathcal A$ be a $\sigma$-algebra of $\Omega$

*$X$ be an $\mathfrak L(E_1,E_2)$-valued random variable on $(\Omega,\mathcal A,\operatorname P)$

*$Y$ be an $E_1$-valued random variable on $(\Omega,\mathcal A,\operatorname P)$


I want to show, that if $X\in\mathcal L^1(\operatorname P,\mathfrak L(E_1,E_2))$, $XY\in\mathcal L^1(\operatorname P,E_2)$ and $Y$ is $\mathcal F$-measurable, then $$\operatorname E\left[XY\mid\mathcal F\right]=\operatorname E\left[X\mid\mathcal F\right]\:Y\;.\tag 1$$
If $E_1=E_2=\mathbb R$, then $(1)$ is an elementary result and I know how to prove it. How can we prove it in the more general case described here?
Besides the proof of $(1)$, I wonder if all the basic properties (like the "tower property", etc.) of the conditional expectation in the real-valued case generalize to the Banach space case. Unfortunately, I couldn't find any textbook which gives more than a simple existence/uniqueness result of the conditional expectation in Banach spaces. So, I would be happy if someone could give me a good reference.
 A: The answer is yes in full generality. You don't need the separability assumption on $E_2$ and the integrability assumption on $XY$ (though clearly you do have to assume that $X\in L^1(\Omega,L(E_1,E_2))$). In comments to the question I was concerned about $E_2$ being a dual space, but that was based on my own confusion and it is irrlevant.
Let me first note that if $\mathcal{F}$ is trivial then $Y$ is simply a vector in $E_1$ and your equation reads 
$$ (*)\quad E[X]Y=E[XY]$$ 
and is straight forward to prove.
For the genral case, I prefer to think of $\mathcal{F}$ as a factor. That is, I consider a measurable map $\pi:\Omega \to \Omega'$ where $(\Omega',\mathcal{A}',P')$ is another probability space in which the points are separated, $\mathcal{F}=\pi^*\mathcal{A}$ and $P'=\pi_*P$. It is known that such a space exists (and it is unique in an apropriate sense)
and for every $\mathcal{F}$-measurable map $Z:\Omega\to E$ there exists an $\mathcal{A}'$-measurable map $Z':\Omega'\to E_1$ such that $Z=Z'\circ\pi$.
We say that $Z$ is the pull-back of $Z'$.
Now we can disintegrate the measure $P$ over $P'$. That is, for a.e $\omega'\in\Omega'$ there exists a probability measure $P_{\omega'}$ on $\Omega_{\omega'}:=\pi^{-1}(\{\omega'\})$ (varying measurably wrt $\mathcal{A}$) such that $P=\int_{\Omega'} P_{\omega'}dP(\omega')$.
For every Banach space $E$ and $\mathcal{A}$-measurable map $Z:\Omega\to E$,
given $\omega'\in\Omega'$ we may consider the restriction $Z_{\omega'}:=Z|_{\Omega_{\omega'}}$ and its expectation wrt $P_{\omega'}$. This defines a function $Z':\Omega'\to E$. It is a basic fact that the conditional expectation $E(Z\mid \mathcal{F})$ is the pull back $Z'$.
The equation $E[XY\mid \mathcal{F}]=E[X\mid\mathcal{F}]Y$ now becomes simply the equation $(*)$, when reduced to the fibers of $\pi$. Indeed, fixing $\omega'\in\Omega'$ and applying $(*)$ for $X_{\omega'}$ and $Y'(\omega')$ (instead of $X$ and $Y$) we get $E[X_{\omega'}]Y(\omega')=E[X_{\omega'}Y(\omega')]$ and pulling back we get the required equation.
Let me conclude by the personal remark that I (a.s) prefer to think geometrically, replacing subalgebras by factors and conditional expectation by integration over fibers. In particular, I prefer "integral" over "expextation". Nevertheless, I am well aware of some advantages of the "probabilistic" mind set over the "geometric" one.
