Let $(\Omega, \mathcal{F}, P)$ be a measure space where $P$ is a probability measure. Let $Q$ be another probability measure that is absolutely continuous with respect to $P$. By Radon-Nikodym, there is $L \in \mathcal{L}^1(P)$ subject to $$ Q(A) = \int_A L(\omega) P(\mathrm{d} \omega) $$ Does it generally hold that $\mathbb{E} \left[L X\right] = \mathbb{E}_Q \left[ X \right]$ for integrable r.v. $X$? I see this result again and again with densities $X = f \cdot m$, but I want to know whether or not this holds without the assumption of having densities.
EDIT: $X, L$ are in $\mathcal{L}^2$.
There is a little bit of ambiguity when you say "$X$ is an integrable random variable." Assuming you mean "$X$ is a $Q$-integrable random variable," yes, this is true. I don't know for sure whether it is true if we only have $X$ is $P$-integrable. To show it is true when $X$ is $Q$-integrable, use the standard method of showing it is true when $X$ is a simple random variable, then show it for general $X$ by approximating with an increasing sequence of simple functions and applying the monotone convergence theorem.