What are properties of $\operatorname E\{X1_A\}$? I have a question in my text about some function $Q : \mathcal A \to \mathbb R$, $Q$(A)= $E\{X1_A\}$, ($X \geq 0$ a.s. and $E\{X\} = 1$).
A later question built on this premise sets $E_Q$ as expectation with respect to $Q$. (Asking to show $E_Q\{Y\}=E_P\{YX\}$.)
Would that mean $E_Q(Y)=E\{E\{X1_Y\}\}$? Or $E_Q(Y)=E\{Y[E\{X1_Y\}]\}$? By the hint from the question, I'm inclined to the latter, but I'm a bit confused.
Also, if I understand correctly, $Q(A)= \sum\{x1_AP(x)\} = \sum_{x\in A} xP(x)$. I haven't reached conditional expectation yet, but this seems like $Q$ could be an expression for that. Am I way off?
 A: You're on the right track, in that conditional expectation indeed deals with expectations of the form $E(XI_A)$. I understand the temptation to try to use conditional expectation. However, in this case, interestingly, one can perfectly answer the question without needing conditional probability.
For simplicity, let's assume a finite possibility space $\Omega$ with probability mass function $p$. Then
$$Q(\{\omega\})=E(X I_{\omega})=X(\omega)p(\omega).$$ So,
$$E_Q(Y)=\sum_{\omega\in\Omega} Q(\{\omega\})Y(\omega)=\sum_{\omega\in\Omega} p(\omega)X(\omega)Y(\omega)=E(XY),$$
establishing the desired equality. All I've used here in the entire argument is the definition of expectation with respect to a probability mass function.
Something not asked but perhaps worth clarifying is that $Q$ actually defines a probability measure on the power set of $\Omega$. To see this, note that

*

*$Q(\{\omega\})=p(\omega)X(\omega)\ge 0$ since $X\ge 0$ a.s. by assumption (which means that $X(\omega)\ge 0$ whenever $p(\omega)\neq 0$, provided $\Omega$ is finite).


*For any $A\subseteq\Omega$ we have that $$Q(A)=E(XI_A)=\sum_{\omega\in A}p(\omega)X(\omega)=\sum_{\omega\in A}Q(\{\omega\})$$ so $Q$ will satisfy the additivity axiom.


*Finally, $$Q(\Omega)=E(XI_{\Omega})=E(X)=1$$ by assumption.
In other words, $Q$ is a non-negative, additive, and normalized set function on the power set of $\Omega$, which means that it is a probability measure.
For the case where $\Omega$ is not finite, some additional details need to be considered, and I'll happily refer you to @Snoop's excellent answer for full technical details.
A: Consider the probability space $(\Omega,\mathcal{A},\mu)$ and the positive random variable $X:\Omega\to \mathbb{R}^+$. Basically, $Q:\mathcal{A}\to [0,1]$ is a probability measure on $(\Omega,\mathcal{A})$. To prove this, notice that
$$Q(\emptyset)=\int\mathbb{I}_\emptyset(\omega)X(\omega)\mu(d\omega)=0$$
$$Q(\Omega)=\int\mathbb{I}_\Omega(\omega)X(\omega)\mu(d\omega)=E[X]=1$$
and for disjoint $(A_n)_{n \in \mathbb{N}}\subset \mathcal{A}$ we have
$$Q(\cup_nA_n)=\int\mathbb{I}_{\cup_nA_n}(\omega)X(\omega)\mu(d\omega)=\int\sum_{n\in \mathbb{N}}\mathbb{I}_{A_n}(\omega)X(\omega)\mu(d\omega)= \\ =\sum_{n\in \mathbb{N}}\int\mathbb{I}_{A_n}(\omega)X(\omega)\mu(d\omega)=\sum_{n\in \mathbb{N}}Q(A_n)$$
where I used monotone convergence to exchange the integral and the summation. In this way we can take expectations using the measure $Q$ for $Y$ which is integrable
$$E_Q[Y]:=\int Y(\omega)Q(d\omega)=\int \lim_{n \to \infty}\sum^{N(n)}_{k=0}\phi_{k,n}\mathbb{I}_{B_{n,k}}(\omega)Q(d\omega)= \\
= \lim_{n \to \infty} \sum^{N(n)}_{k=0}\phi_{k,n}Q(B_{n,k})= \lim_{n \to \infty} \sum^{N(n)}_{k=0}\phi_{k,n}\int \mathbb{I}_{B_{n,k}}(\omega)X(\omega)\mu(d\omega)= \\ = \int \lim_{n \to \infty}\sum^{N(n)}_{k=0}\phi_{k,n}\mathbb{I}_{B_{n,k}}(\omega)X(\omega)\mu(d\omega)=\int Y(\omega)X(\omega)\mu(d\omega)= E[YX]$$
where I used a sequence of simple functions approximating $Y$ and dominated convergence to take the limits in and out the integral.
