5
$\begingroup$

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?

$\endgroup$
4
  • $\begingroup$ The proof is correct, but you might write standard Borel space instead of just Borel space. The latter is Mackey's term for a measurable space. $\endgroup$ Jul 11, 2013 at 13:07
  • $\begingroup$ @MichaelGreinecker: thanks, I just followed the terminology of the Bertsekas and Shreve's book. Have you seen such a statement as in OP somewhere? I haven't find it in BS. $\endgroup$
    – SBF
    Jul 11, 2013 at 13:13
  • 1
    $\begingroup$ Lemma 1.2 in Crauel, Random Probability Measures on Polish Spaces, is slightly stronger. It says that if a function from a probbility space to a separable metric space is measurable with respect to the completion, it equal some measurable function almost surely. $\endgroup$ Jul 11, 2013 at 13:22
  • $\begingroup$ @MichaelGreinecker: nice, would you post this as an answer? $\endgroup$
    – SBF
    Jul 11, 2013 at 13:24

1 Answer 1

5
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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