Orthogonality of joint probability and conditional probability measures

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Suppose that $(X,Y)$ are real valued random variables on some space $(\Omega, \mathcal{F})$. Let $P,Q$ be two possible joint probability measures for $(X,Y)$. Let $P_{Y|X}$ and $Q_{Y|X}$ be two possible regular conditional probability measures for $Y|X$. I understand the following definition:

Two probability measures $P$ and $Q$ are orthogonal (denoted as $P \perp Q$) if there exists $A \in \mathcal{F}$ such that $P(A) = 0$ and $Q(A) = 1$.

Does the orthogonality of $P,Q$ relate in any way to the orthogonality of $P_{Y|X}$ and $Q_{Y|X}$? My guess is that $P \perp Q$ implies $P_{Y|X} \perp Q_{Y|X}$ since we seem to be restricting $P,Q$ to the smaller $\sigma$-algebra $\sigma(X)$ in this case. How about the converse? Is this question even well defined, because we should really be considering $P_{Y|X = x}$ and $Q_{Y|X = x}$ for a given $x \in \mathbb{R}$.

My question is motivated by these lecture notes on information theory. Particularly, the part where they start talking about conditional divergence and entropy. I see expressions like $\frac{P_{Y|X}}{Q_{Y|X}}$ used, which I'm guessing represents a Radon-Nikodym derivative. But since we're conditioning, I'm not sure what this means exactly. I don't think there's a "conditional" version of the Lebesgue decomposition out there.

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$P$ and $Q$ being orthogonal means they have disjoint supports (a set $A$ is said to be a support of a measure $P$ if $P(A^c) = 0$). Suppose $P$ is supported on $A$ and $Q$ is supported on $B$. Let $x \in \mathbb{R}$. $P_{Y \mid X = x}$ is supported on $\{y : (x, y) \in A\} = A_x$ and $Q_{Y \mid X = x}$ is supported on $B_x$. So your question is whether $A \cap B = \emptyset$ implies $A_x \cap B_x = \emptyset$. This is true since $(A \cap B)_x = A_x \cap B_x$.

Technically, $P_{Y \mid X}$ is a stochastic kernel, which means that for each $x \in \mathbb{R}$, $P_{Y \mid X = x}$ is a probability measure, and that for every $A \in \mathcal{B}(\mathbb{R})$, the map $x \mapsto P(A)_{Y \mid X = x}$ is measurable. So this means that $P_{Y \mid X = x}$ is a measure, and all theorems of measure theory apply to it.