If $X$ is $\mathcal G$-measurable and $Y$ is independent of $\mathcal G$, then $\text E[XY\mid\mathcal F]=\text E[Y]\text E[X\mid\mathcal F]$

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
• $\mathcal F,\mathcal G$ be $\sigma$-algebras of $\Omega$ with $\mathcal F\subseteq\mathcal G\subseteq\mathcal A$
• $X$ 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 $X$ is $\mathcal G$-measurable and $Y$ is independent of $\mathcal G$, then $$\operatorname E\left[XY\mid\mathcal F\right]=\operatorname E\left[X\mid\mathcal F\right]\operatorname E\left[Y\right]\;.\tag 1$$ $(1)$ is easy to prove in the case of real-valued random variables $X,Y$. How can we show it in the given case?

• To be clear, I know how I can prove the statement in the case of real-valued $X,Y$, but it's not clear to me how the necessary properties of the conditional expectation generalize to the case described in the question. – 0xbadf00d Oct 6 '16 at 20:06
• I guess the downvoters didn't read the question and didn't understand why I ask for the proof of an elementary result. – 0xbadf00d Oct 6 '16 at 21:49