# Is $(X, Y)$ always absolutely continuous with respect to $P_X \otimes P_Y$?

Definitions:

Let $$X: (\Omega, \mathcal A) \to (\mathbb R, \mathcal B)$$ be a random variable on the probability space $$(\Omega, \mathcal A, P)$$ and define its distribution as the probability measure $$P_X(B) = P(X \in B)$$ on $$\mathcal B$$. ($$\mathcal B$$ is the Borel sigma algebra).

A random variable is absolutely continuous with respect to a measure $$\mu$$ if its distribution is, i.e. $$P_X(B)=0$$ for all $$B \in \mathcal B$$ with $$\mu(B) = 0.$$ $$X$$ may not be absolutely continuous with respect to the Lebesgue measure $$\lambda$$ but it is always absolutely continuous with respect to $$P_X$$.

Now let $$(X,Y): (\Omega, \mathcal A) \to (\mathbb R^2, \mathcal B^2)$$ be a random vector with distribution $$P_{X, Y}((X, Y) \in B)$$, $$B \in \mathcal B^2$$. By the same reasoning as before, $$(X,Y)$$ is abolutely continous with respect to $$P_{X, Y}$$ even if it is not absolutely continuous with respect to $$\lambda^2$$.

Question: Will $$(X, Y)$$ always be absolutely continuous with respect to the product measure $$P_X \otimes P_Y$$?

What I did:

We need to verify that $$P_{X, Y}(B) = 0$$ whenever $$(P_X \otimes P_Y)(B)=0$$. If $$B = B_1 \times B_2$$ is a rectangular set then this is clearly true because $$(P_X \otimes P_Y)(B)=0$$ implies $$P_X(B_1) = 0$$ or $$P_Y(B_2)=0$$ ($$P_X(B_1) = 0$$, say) and then $$P_{X, Y}(B) = P((X, Y) \in B_1 \times B_2) = P(X \in B_1, Y \in B_2) \le P(X \in B_1) = 0.$$

But for non-rectangular sets I'm not sure how to proceed.

This is untrue in general. The problem is that $$P_X \otimes P_Y$$ is the law of $$(\tilde{X}, \tilde{Y})$$ where $$\tilde{X}$$ has the same distribution as $$X$$ and similarly for $$\tilde{Y}$$ but $$\tilde{X}$$ is independent from $$\tilde{Y}$$. $$X$$ and $$Y$$ need not be independent which will allow us to create a counterexample.
For example, let $$X$$ be a standard one dimensional Gaussian and consider the case $$X = Y$$. Then $$P_{(X,X)}$$ is a measure on $$\mathbb{R}^2$$ such that if $$\Delta = \{(x,x) \in \mathbb{R}^2: x \in \mathbb{R}\}$$ then $$P_{(X,X)}(\Delta) = 1$$. However, $$P_X \otimes P_X$$ is the law of the standard two dimensional Gaussian and in particular $$P_X \otimes P_X(\Delta) = 0$$.