Let $X$ be a standard Borel space: that is, a topological space equivalent to a Borel subset of $\Bbb R$. It is known that for any probability measure $p$ on $X$ and any universally measurable set $A\subseteq X$ there exists a Borel set $B\subseteq X$ such that $p(A\Delta B) = 0$. Moreover, if $f:X\to\Bbb R$ is universally measurable, there exists a Borel-measurable $f'$ such that $f=f'$ $p$-a.e. 


I wonder, whether the following result holds true:



> 
> If $Y$ is also a Borel space and $g:X\to Y$ is universally measurable, there exists a Borel map $g':X\to Y$ such that $g = g'$ $p$-a.e.
> 
> 
> 


A possible proof that I have in my mind is the following: the countable case is trivial, so assume that $Y$ is uncountable, then there exists a Borel isomorphism $\phi:Y\to\Bbb R$. As a result, we can say that
$$
 g = \phi^{-1}\circ (\phi\circ g)
$$
where $\phi\circ g:X\to\Bbb R$ is clearly universally measurable. Hence, there exists a Borel function $f':X\to\Bbb R$ such that $\phi\circ g = f'$ $p$-a.e. If we define $g':=\phi^{-1}\circ f'$, we obtain that $g':X\to Y$ is a Borel map and that $g=g'$ $p$-a.e. 


Is the proof correct?



Universally measurable map coincides a.e. with a Borel map
The proof is correct. The following result of a stronger assertion is taken from Hans Crauel's book *Random Probability Measures on Polish Spaces*, where it is Lemma 1.2.:


**Proposition:** Let $(\Omega,\Sigma,\mu)$ be a probability space, $Y$ a separable metric space and $f\_0:\Omega\to Y$ be measurable with respect to the $\mu$-completion of $\Sigma$. Then there is a measurable function $f:\Omega\to Y$ such that the set $\{\omega:f\_0(\omega)\neq f(\omega)\}$ has $\mu$-outer measure zero.


