Let $X$ be a random variable on a probability space $(\Omega,\mathscr F, P)$. Define a new probability measure $$\tilde P(A) = E[1_A X]$$ for all $A\in\mathscr F$. Let $\tilde E$ be expectation taken with respect to the new measure $\tilde{P}$.
Suppose now that $Y$ is also a random variable $(\Omega,\mathscr F)$. Then intuitively the expectation should be computed as $$ \tilde E [1_A Y] = E[1_A YX], $$ but I'm not sure how to prove this rigorously using the definition.
On
I'm changing $\tilde{\mathbb P}$ to $\mathbb Q$.
We're given that
$$\mathbb Q(A) = E^{\mathbb P}[X1_A] \tag{1}$$
$$LHS(1) := \mathbb Q(A) = \int_{\Omega}1_A d\mathbb Q$$
$$RHS(1) := \int_{\Omega} X1_A d\mathbb P$$
Thus, $$X1_Ad\mathbb P = 1_Ad\mathbb Q$$
which I guess is equivalent to
$$Xd\mathbb P = d\mathbb Q \tag{*}$$
Now, we want to prove that
$$E^{\mathbb Q}[1_AY] = E^{\mathbb P}[1_AXY] \tag{2}$$
$$LHS(2) := \int_{\Omega} Y1_A d\mathbb Q$$
$$RHS(2) := \int_{\Omega} XY1_A d\mathbb P$$
Now, $LHS(2) = RHS(2)$ if $XY1_Ad\mathbb P = Y1_Ad\mathbb Q$
which I guess is equivalent to $(*)$.
This is a standard exercise in proving identities with the Lebesgue integral. Prove it first with a simple function, i.e. if $\phi(\omega) = \sum a_i \chi_{A_i}(\omega)$ then $$ \int_A \phi(\omega) \tilde{P}(d\omega) = \int_A \phi(\omega)X(\omega) P(d\omega). $$ Now prove it for non-negative random variables. To do this, it should be known that there is a sequence of increasing simple functions $\phi_n \to Y$, and apply the previous result and monotone convergence theorem. Then the result follows for random variables follows.