Is every probability measure a pushforward of Lebesgue measure?

Question

If $m$ is a probability measure on a measurable space $(X, \Sigma)$, is there necessarily a measurable function $f : [0, 1] \to X$ such that $m(A) = \mu(f^{-1}(A))$ for all $A \in \Sigma$? ($\mu$ is the Lebesgue measure on $[0,1]$)

Answer 1

No. Some probability spaces are too big. An example should be $X = \{0,1\}^A$, with the Haar measure, where $A$ has large enough cardinal.

Probably $|A| = 2^{\aleph_0}$ would do it, but the proof for that would require some work.

But let's do it without that much work. If we make the cardinal of $A$ really big, we can get: for any set $E \subset X$ of cardinal $2^{\aleph_0}$, the closure of $E$ has measure zero.

added
A simpler example with the same idea.
Let $X$ be a set with cardinal ${}\gt 2^{\aleph_0}$. Let the sigma algebra $\Sigma$ consist of all subsets of cardinal $\le 2^{\aleph_0}$ and their complements. The measure $m$ is defined as: $m(E) = 0$ for sets $E$ with cardinal $\le 2^{\aleph_0}$, and $m(X\setminus E) = 1$ for their complements.

I claim there is no map $f : [0,1] \to X$ with the condition required. Indeed, let $f : [0,1] \to X$ be any map. The image $A = f\big([0,1]\big)$ has cardinal $\le 2^{\aleph_0}$, so $m(A) = 0$, but $\mu\big(f^{-1}(A)\big) = \mu\big([0,1]\big) = 1$.

Answer 2

There are even counterexamples with $X=[0,1]$ and $\Sigma$ countably generated. Gnedenko and Kolmogorov introduced the notion of a perfect probability measure. One characterization of perfectness is that the probability space $(X,\mathcal{X},\nu)$ is perfect if whenever $f:X\to\mathbb{R}$ is measurable, there must be a Borel set $B\subseteq f(X)$ such that $\nu\circ f^{-1}(B)=1$. The pushforward of a perfect probability measure is then clearly perfect again and it follows from the inner regularity of Lebesgue measure that Lebesgue measure on $[0,1]$ is perfect. So it suffices to give a probability measure that is not perfect. Now perfect probability measures are very well-behaved. In particular, for perfect probability spaces, regular conditional probabilities with respect to a countably generated sub-$\sigma$-algebra always exist. But most advanced probability theory textbooks will give you an example of a probability measure on $[0,1]$ with the $\sigma$-algebra constructed from the Borel sets and one nonmesuarable set, such that no regular conditional probability with respect to the Borel sets exists.

Answer 3

(I know next to nothing about exotic measure spaces, but here is what I was able to find).

This question is very close to Maharam's theorem, which asserts that every complete measure space is "isomorphic" to a weighted sum of products of the standard measure space ($[0,1]$ with the Lebesgue measure). Important note: "isomorphic" refers to algebras of measurable sets modulo null sets, and not the underlying spaces (see here for a detailed discussion).

If $|X| \leqslant 2^{\aleph_0}$, then necessarily $X$ is "isomorphic" to $[0,1]$ plus atoms. I do not know, however, if this "isomorphism" can be lifted to a point-wise isomorphism, which would clearly imply a positive answer to the original question. A perforated interval might be a counterexample here, but straight away I do not see why. Actually, the entire Wikipedia article on standard probability spaces might give a good start.

Answer 4

Lebesgue measure on $[0, 1]$ has non separable extensions. See

A Non-Separable Translation Invariant Extension of the Lebesgue Measure Space, Kunihiko Kodaira and Shizuo Kakutani, Annals of Mathematics Second Series, Vol. 52, No. 3 (Nov., 1950), pp. 574-579

This gives a negative answer with $X = [0, 1]$.