Suppose $X$ and $Y$ are random variables, and let $P_Y$, $P_X$, and $P_{Y \mid X = x}$ denote the probability distribution of $Y$, the probability distribution of $X$, and the conditional probability distribution of $Y$ given $X = x$. Let $f$ be a function whose domain is the support of the random variable $Y$. Let a.e. and a.s. denote abbreviations of almost everywhere and almost surely.
Question: Is $f$ is continuous at $P_Y$-a.e. $y$ if and only if it is continuous at $P_{Y \mid X = x}$-a.e. $y$ for $P_X$-a.e. $x$?
More generally, is it correct to say that some property $V$ of the random variable $Y$ holds $P_Y$-a.s. if and only if it holds $P_{Y \mid X = x}$-a.s. for $P_X$-a.e. $x$?
I can believe that if property $V$ holds $P_{Y \mid X = x}$-a.s. for $P_X$-a.e. $x$, then it is possibly holds $P_Y$-a.s. since this seems similar to the result that a countable union of measure zero sets is measure zero (I don't have a proof, though). I am not even sure that the converse is true.
Best Answer
The core of your question does not seem to be related to the function $f$ at all. If I understand you correctly, your problem is solved by the continuous version of the law of total probability (compare this question), which says that
$$ P_Y(B)=\int_{\Bbb R^d} P_{Y|X=x}(B) P_X(dx)$$
for every measurable set $B\subset \Bbb R^d$ (sticking to the euclidean case for simplicity). This integral is equal to $1$ if and only if the integrand is equal to $1$ for $P_X$-almost all $x\in \Bbb R^d$ (since the integrand is $[0,1]$-valued and $P_X$ is a probability measure). In other words, something indeed happens $P_Y$-almost surely if and only if it happens $P_{Y|X=x}$-almost surely for $P_X$-almost all $x\in \Bbb R^d$.