If $Q \ll P$ and $X$ is $P$-integrable, does $\int X dQ$ exist?

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The title says it all. I need to show that not both the positive and negative parts of $X$ have infinite expectation.

I know that the integrability assumption implies $P(|X|>n) \to 0$ and so $Q(|X|>n) \to 0$ by absolute continuity. But that’s not enough to conclude anything.

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The unswer to the topic question is 'No, in general'.

Probability measure $Q$ is absolute continuous with respect to $P$, if for every $P$-measurable $A$, $P(A)=0$ implies $Q(A)=0$. Say, let $P$ to be the exponential distribution with mean $1$ and let $Q$ to be Pareto distribution with PDF $f(x)=\frac{1}{x^2}\mathbb 1_{\{x\geq 1\}}$. Then $Q\ll P$.

These probability measures are defined on the probability space $$ \Omega=\mathbb R, \quad \mathcal F = \mathfrak B(\mathbb R). $$

Consider random variable $X(\omega)=\omega$. Then the distribution of $X$ is exponential under $P$ and Pareto under $Q$, and the expectation of $X$ exists under $P$ and does not exists under $Q$.