# Is a conditional probability measure dominated by its unconditional counterpart?

A dilettante here trying to better understand the basic definition/properties of conditional probabilities.

In my setting I have a probability triple $$(S, \Sigma, P)$$ and a random vector $$(X,Y)$$ living on that space and mapping to a measurable space with a product space structure $$(\textbf{X \times Y}, \mathcal{X} \otimes \mathcal{Y})$$.

Let $$\mu$$ be the "law"/"push-forward measure" associated with $$(X,Y)$$ and let $$\mu_X$$, $$\mu_Y$$ be the (marginal) laws of $$X$$ and $$Y$$ defined through "projection", e.g. $$\mu_X(B) = \mu(B\times \textbf{Y})$$ for $$B \in \mathcal{X}$$. Suppose conditional probability measures denoted by $$\mu_{X | Y=y}$$ are well-defined (regular) for $$y \in \textbf{Y}$$.

My question: Is it true that $$\mu_{X | Y=y}$$ is absolutely continuous w.r.t. $$\mu_X$$ for almost all $$y$$? Intuitively, it would make sense to me that if X being in some range is "impossible", getting more information through $$Y$$ cannot suddenly make this "possible".

I have been attempting a contradiction argument and assume there is a set $$A \in \mathcal{Y}$$ with $$\mu_Y(A) >0$$ s.t. for all $$y \in A$$ there is a $$B_y$$ with $$\mu_X(B_y) = 0$$ but $$\mu_{X|Y=y}(B_y) >0$$. I have tried to get a contradiction by using the definition of conditional probabilities, in particular that for $$C \in \mathcal{X}$$ we have $$\int_A \mu_{X | Y=y}(C)dy = \mu(C \times A)$$, but this has not been fruitful for me so far.

This is not true. Consider $$(X, Y) = (Z,Z) \in \mathbb{R}^2$$ where $$Z$$ is a $$\mathcal{N}(0,1)$$ random variable. Then $$\mu_X = \mathcal{N}(0,1)$$ but $$\mu_{X \mid Y=y} = \delta_{y}$$, the Dirac measure at point $$y$$.