About conditional expectation and R-N derivative: $X \geq 0$ is a r.v. On $(\Omega, \mathcal{F}, \mathbb{P})$. $\mu(E)=\mathbb{E}(X\chi_{E})$

Question

$X \geq 0$ is a random variable on $(\Omega, \mathcal{F}, \mathbb{P})$, $\mathbb{E}(X) = 1$.

$\mathcal{G} \subset \mathcal{F}$ is a sub sigma-algebra. Define $\mu(E)=\mathbb{E}(X1_{E})$.

How to show $\mu$ is absolutely continuous w.r.t. $\mathbb{P}_{|\mathcal{G}}$ and find the Radon-Nikodym derivative?

Edit: Really appreciate @John Dawkins for pointing out that the bar symbol means the restriction, not the "conditioning on"

Answer 1

This is more of an extended comment than a full answer:

If $E\in\mathcal G$ with $\mathbb P(E\mid\mathcal G)=0$ then $$\mu(E)=\mathbb E[X\mathsf 1_E\mid\mathcal G]\leqslant \mathbb E[|X|\mathsf 1_E\mid\mathcal G]\leqslant \mathbb E[|X|\mid\mathcal G]\mathbb P(E\mid\mathcal G)=0.$$ It seems that the Radon-Nikodym derivative should be: $$ \mu(E) = \int_E X\ \mathsf d \mathbb P_{\mid\mathcal G}. $$

Answer 2

Suppose that $E\in\mathcal G$ and $\Bbb P_{|\mathcal G}(E) =0$. This latter condition simply means that $\Bbb P(E)=0$. Therefore $1_E=0$, $\Bbb P$-a.e. Consequently $X1_E=0$, $\Bbb P$-a.e., as well, and $$ \mu(E) =\Bbb E[X1_E] =0. $$ This proves the implication $\Bbb P_{|\mathcal G}(E) =0\,\Rightarrow \mu(E)=0$, for each $E\in\mathcal G$. Thus $\mu\ll \Bbb P_{|\mathcal G}$.

[N.B. $\Bbb P_{|\mathcal G}$ is the restriction of $\Bbb P$ to $\mathcal G$; it is not the conditional probability $\Bbb P[\cdot\mid\mathcal G]$.]