Assume that we have a probability space $(\Omega, \mathcal{A}, P)$, and we have two random variables $X,Y: \Omega \rightarrow \mathbb{R}$. On this space.
We can define two measures $\mu_X(B)=P(X^{-1}(B)), \mu_Y(B)=P(Y^{-1}(B)), B \in \mathcal{B}(\mathbb{R})$.
If $\mu_X$ and $\mu_{Y}$ is absolutely continuous with respect to the Lebesgue measure we get that $\mu_X(B)=\int_Bf_Xd\lambda,\mu_Y(B)=\int_Bf_Xd\lambda$, where $\lambda$ is the Lebesgue measure. f is called the probability density. (This follows from the Radon-Nikodym Theorem).
We also have that $\mu_{XY}(B_2)=P((X,Y)\in B_2), B_2\in \mathcal{B}(\mathbb{R}^2)$.
My question is this:
If we know that $\mu_X, \mu_Y$ is absolutely continuous with the respect measure, can we then deduce that $\mu_{XY}$ is absolutely continuous with respect to the Lebesgue measure on $\mathbb{R}^2$, and then there exists a probability density function such that $\mu_{XY}(B_2)=\int_{B_2}f_{XY}(r,t)d(\lambda\times \lambda)?$
I am not able to prove this. I am able to prove the converse that if $\mu_{XY}$ is absolutely continuous with respect to the Lebesgue measure on $\mathbb{R^2}$ we also have absolute continuity in separate variables, is this correct? Could you please tell me which directions are correct, and which do not hold?