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.

  • $\begingroup$ It is unclear to me what the domain of $f$ is. $\endgroup$ Jul 31, 2019 at 18:01
  • $\begingroup$ @GiuseppeNegro Suppose the random variable $Y$ has support $\Pi \subset \mathbb{R}^d$ (we assume Euclidean space for simplicity). Then, we can assume that $f$ has domain $\Pi$. $\endgroup$ Jul 31, 2019 at 18:03

1 Answer 1


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 $x\in \Bbb R^d$.

  • 1
    $\begingroup$ Thanks, it looks like this link states a continuous version of the law of total probability. I agree that the function $f$ is irrelevant here, which means that the more general assertion is true under measurability assumptions. $\endgroup$ Jul 31, 2019 at 19:22
  • $\begingroup$ Good answer. Thanks for writing it. $\endgroup$ Aug 9, 2019 at 20:07

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