MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

It is standard that every Borel probability measure on a polish space $X$ can be obtained as pushforward of the uniform measure $\lambda$ on $[0,1]$ along an almost-everywhere-defined Borel-measurable function $d: [0,1] \to X$ . (In fact, $d$ can always be taken to be continuous on a measure-$1$ $G_\delta$ subset. But this is not important for what follows.) My question is whether there is a "conditional" version of this result, along the following lines. I first formulate this below as a precise question, and then discuss possible variations.

I shall state the question as a property of the category whose objects are polish probability spaces (i.e., a polish space together with a Borel probability measure) and whose maps are almost-everywhere-defined measure-preserving Borel-measurable functions modulo almost-everywhere equality. Note that, in terms of this category, the fact stated at the start asserts that $[0,1]$ (with $\lambda$ as its probability measure) is a weakly initial object (i.e., there exists at least one map from $[0,1]$ to every object $X$.)

Precise question. Given a map $f: X \to Y$, does there exist a map $e : Y \times [0,1] \to X$ such that $f \circ e = \pi_1$ (where $\pi_1$ is first projection)? (Here, $Y \times [0,1]$ is the topological product with product Borel measure.)

Remarks and variations

  1. This is a conditional version of the initial fact in the following sense. Given such an $e$, the function mapping $y \in Y$ to the pushforward of $\lambda$ along $e(y,-): [0,1] \to X$ is a disintegration of $f$, thus giving conditional probability measures $P(-|f(x) = y)$.

  2. Category-theoretically, the question asks if every projection $\pi_1: Y \times [0,1] \to Y$ is a weakly initial object in the slice category over $Y$.

  3. For the application I have in mind, a weaker result would suffice. It is enough for there to exist $d: [0,1] \to Y$ and $e:[0,1] \times [0,1] \to X$ such that $f \circ e = d \circ \pi_1$.

  4. I am not that fussy about the precise choice of category. One might, e.g., generalise to analytic or standard probability spaces. In general, I'm interested in pointers to any result at all that is similar in nature to the one discussed.

share|cite|improve this question
This sounds a lot like something that's done in the category of "Lebesgue spaces", a Lebesgue space being defined axiomatically by the existence of a sequence of finite partitions that separates points of the space. In this category, I think the answer is positive. Details can be found in the first chapter of the book by Rudolph: Fundamentals of Measurable Dynamics – Anthony Quas Apr 17 '13 at 22:55
up vote 4 down vote accepted

It follows from a classification of morphisms in this category due to Rokhlin. If both the target space and all the conditional measures are purely non-atomic, then this map is (mod 0) just the coordinate projection of the unit square (endowed with the Lebesgue measure). If any of these measures has atoms, then it is essentially the same description with obvious modifications.

share|cite|improve this answer
Thanks very much. Indeed, this is to be found in Section 4 of Rohlin's "On the fundamental ideas of measure theory" paper. – Alex Simpson Apr 18 '13 at 9:28

I'am not sure this is what you want, but there is a way to represent regular conditional probabilities as some kind of pushforward-measure. Let $\kappa$ be a probability kernel from $Y$ to $X$. That is $\kappa:Y\times\mathcal{B}(X)\to[0,1]$ is measurable as a function of $Y$ and a probability measure as a function of $\mathcal{B}(X)$. Then there is a measurable function $f:Y\times[0,1]\to X$ such that $\kappa(y,\cdot)$ is the distribution of $f(y,\cdot)$. This can be found as Proposition 10.7.6 in Bogachev's Measure Theory.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.