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Let $X: \Omega\to{\mathbb R}$ be a random variable. Is it always possible to modify it (i.e. change the value of $X$ on a subset of $\Omega$ of zero measure) so that the range of $X$ is a Borel set?

This is related to the following question: we know that $E(Y|X)$ can be written as a function of $X$, i.e. $E(Y|X)=\varphi(X)$ for some $\varphi: {\mathbb R}\to{\mathbb R}$. Is $\varphi$ (Borel) measurable? We can prove this if we know $X$ has Borel range.

I guess the answers might be negative, will there be (easy) counterexamples? Can we add some assumptions to make the conclusion true?

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    $\begingroup$ If you are interested in probability you may assume that the measure is complete (each subset of a set of measure zero is measurable). In many cases $\Omega$ will contain a set of measure zero with cardinal the continuum. Then you may modify the random variable in this measure zero set so that the range is $\Bbb R$. $\endgroup$
    – juan
    Commented May 10, 2021 at 21:31
  • $\begingroup$ The function $\varphi$ in your question is not unique (since the equality $E(Y \mid X)=\varphi(X)$ can only be expected to hold a.s.), and it is indeed true that you can always choose it to be Borel measurable. That should be part of the statement - it would be a pretty useless theorem if this were not true. $\endgroup$
    – Nate Eldredge
    Commented May 10, 2021 at 23:25
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    $\begingroup$ The related result is true even without the need to modify the conditional expectation (i. e., for any choice of one), see e. g. Lemma 1.13 in Kallenberg, Foundations of modern probability. $\endgroup$
    – Kostya_I
    Commented May 11, 2021 at 7:18
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    $\begingroup$ The relevant condition on a probability measure is being perfect, a notion due to Gnedenko and Kolmogorov. $\endgroup$
    – Michael Greinecker
    Commented May 12, 2021 at 22:57

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Assume that $(\Omega, {\mathcal F}, P)$ itself is a Lebesgue space, so it can be realized as a Polish space equipped with the completion of the Borel sets. If $X$ is a random variable, it can be changed on a set of measure zero to be a Borel measurable function. (This is an easy consequence of [1]). Thus we may assume $X$ itself is Borel measurable.

The class of analytic sets is closed under direct images by Borel functions, see e.g. Proposition 1.4 in [2], or [3]. Finally, any analytic set is universaly measurable [3], so $X(\Omega)$ differs from a Borel set $B$ on a set $A$ of zero measure with respect to the push-forward $\mu:=PX^{-1}$, that is $B=X(\Omega) \setminus A$. Thus if $X^*$ is obtained from $X$ by changing it to map $X^{-1}(A)$ to one point in $B$, then $X^*(\Omega)=B$.

[1] https://en.wikipedia.org/wiki/Monotone_class_theorem

[2] https://webusers.imj-prg.fr/~dominique.lecomte/Chapitres/6-Analytic%20and%20co-analytic%20sets.pdf

[3] Kechris, A. Descriptive set theory, Springer.

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