My book claims that if $Z$ is a random variable and $Y$ is $\sigma(X)$-measurable, then $Y$ is a function of $X$ and does not depend on $\omega \in \Omega$. I can't follow the proof in the book.

If $Z$ is $\sigma(X)$-measurable, then for each $z\in\mathbb{R}$, $\{Z = z\} = \{X \in B_z\}$ for some Borel $B_z \subset \mathbb{R}$ (make sense to me). Then $Z = f(X)$ where $f$ is defined by $f(x) =z$ for all $x\in B_z$ (Huh?).

The second sentence does not make sense to me. For example, what if $x\notin B_z$?

I tried to write my own proof but had some problem in the last step: Let $\omega_1 \ne \omega_2$ and $X(\omega_1) = X(\omega_2)$. We want to show that $Z(\omega_1)=Z(\omega_2)$. For any Borel set $B \subset \mathbb{R}$, we consider the set $\{X \in B\}$. We either have both $\omega_1$ and $\omega_2$ are in $\{X \in B\}$, or we have both $\omega_1$ and $\omega_2$ are not in $\{X \in B\}$.

In particular, for $B_z$, if both $\omega_1$ and $\omega_2$ are in $B_z$, then $Z(\omega_1 ) = Z(\omega_2) = z$. If $\omega_1$ and $\omega_2$ are not in $B_z$, then I got stuck.

  • $\begingroup$ @SangchulLee Oops, typo. $\endgroup$
    – 3x89g2
    Oct 12, 2017 at 15:36
  • $\begingroup$ Well, even though the highlighted argument can be made legit, it still does not provide any useful information unless we establish some measurability on $f$. A standard argument uses the machinery called monotone class theorem, and the result ensures that $Z = f(X)$ for some Borel measurable $f : \mathbb{R} \to \mathbb{R}$. $\endgroup$ Oct 12, 2017 at 15:39
  • $\begingroup$ @SangchulLee To provide some context, we are discussing the definition of $E(Y\mid X)$ as a random variable. The definition requires that $E(Y\mid X)$ to be $\sigma(X)$-measurable. The quote talks about why we require that. The author did not mention anything about the theorem that you mentioned. $\endgroup$
    – 3x89g2
    Oct 12, 2017 at 15:56
  • $\begingroup$ What is your source? The approach you explain in your post is the naïve one, which works for discrete random variables and fails utterly in the general case. $\endgroup$
    – Did
    Oct 12, 2017 at 16:30
  • $\begingroup$ @Did I am reading First Look at Rigorous Probability Theory by Jeffrey S. Rosenthal $\endgroup$
    – 3x89g2
    Oct 12, 2017 at 16:40

2 Answers 2


Regarding your remark "For example, what if $x\notin B_z$?": The author seems to be defining a function $f:\cup_{z\in\mathbb R} B_z\to\mathbb R$ as you described. So if $x\in\cup_{z\in\mathbb R} B_z$, $x\in B_z$ for some $z$, in which case we define $f(x)=z$.


If $X(\omega_1)=X(\omega_2)$ and $Z(\omega_1)\ne Z(\omega_2)$, then there is a Borel set $B$ s.t. $Z(\omega_1)\in B$ and $Z(\omega_2)\notin B$. Since $Z$ is $\sigma(X)$-measurable, $Z^{-1}(B)=X^{-1}(A)$ for some Borel set $A$ s.t. $X(\omega_1)\in A$ and $X(\omega_2)\notin A$, which is a contradiction.

This gives a measurable function $f:\mathcal{X}\to\mathbb{R}$, where $\mathcal{X}$ is the range of $X$, s.t. $Z(\omega)=f(X(\omega))$.

  • $\begingroup$ How does this specify a measurable $f$? $\endgroup$ Oct 15, 2017 at 1:00
  • $\begingroup$ @JohnDawkins This does not specify $f$. It implies that $Z=f\circ X$ for some function $f$ on the range of $X$. $f$ is $\mathcal{B}_{\mathcal{X}}$-measurable because for any $B\in \mathcal{B}_{\mathbb{R}}$, $f^{-1}(B)=\mathcal{X}\cap A$ for some $A\in\mathcal{B}_{\mathbb{R}}$. This $f$ can be extended to $\mathbb{R}$ by usual approximation argument. $\endgroup$
    – user140541
    Oct 15, 2017 at 4:32
  • $\begingroup$ The function $f$ is uniquely determined on $\mathcal X$ by $f(x) =z$ for all $x\in Z^{-1}(\{z\})$. But (to play the Devil's Advocate) how does this yield the measurability of $f$ (i.e. the assertion that $f^{-1}(B)=\mathcal X\cap A$ with $A\in\mathcal B_{\Bbb R}$, for each Borel set $B$), unless the range of $Z$ is countable? $\endgroup$ Oct 15, 2017 at 15:24
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    $\begingroup$ My mistake. I lost track of the fact that you were only looking to define a function $f$ on the range of $X$, and not (as usual) to find Borel $f:\Bbb R\to\Bbb R$ with $Z=f\circ X$. $\endgroup$ Oct 16, 2017 at 19:53
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    $\begingroup$ If you are going to do that, then you may as well prove the entire result as the limit of simple functions. =) (en.m.wikipedia.org/wiki/Doob%E2%80%93Dynkin_lemma) I believe the extension I gave in my previous comment works and is a bit more explicit (in particular, we only need to worry about defining $f(x)$ on the boundary of $\mathcal{X}$). Of course it needs $f$ to be bounded. $\endgroup$
    – Michael
    Jul 17, 2021 at 21:51

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