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 $[0,1]$)



Is every probability measure a pushforward of Lebesgue measure?
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 $\mu\big(f^{-1}(A)\big) = \mu\big([0,1]\big) = 1$.


