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If $f,g:(\Omega, \mathcal{F},P) \to (\mathbb{R}, \mathcal{B}(\mathbb{R}))$ are measurable, and $f=h\circ g$ where $h: \mathbb{R}\to \mathbb{R}$, then is $h$ Borel-measurable?

The above question is motivated by the following statement: for any $x,y\in \Omega$ if $g(x) = g(x)$ implies that $f(x)=f(y)$, then $f$ is a function of $g$, say $f = h\circ g$. Here, how can we analyze the measurability of $h$? Alternatively, can we claim that there is a $\tilde h$ such that $\tilde h = h$ almost everywhere?

(I tried to approach this using: for any set with positive outer measure, there is a non-measurable subset $A$ of it, therefore $\mathbb{1}_{A}$ is non-measurable. Yet it is hard to characterize a non-measurable function in general.)

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  • $\begingroup$ Borel-measurable means that pre-image of Borel set is Borel. Assume $X$ is Borel, then $f^{-1}(X) = g^{-1}(h^{-1}(X))$, $h$ is measurable so $h^{-1}(X)$ is Borel, and as $g$ is measurable, $g^{-1}(h^{-1}(X))$ is also Borel. $\endgroup$
    – mihaild
    Commented Sep 25, 2022 at 20:55
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    $\begingroup$ Thanks for your answer. Yet I assumed the measurability of $f,g$ here and want to find out the measurability of $h$. The statement you proposed shows that given $g$ measurable and $h$ Borel-measurable, their composition is measurable. $\endgroup$ Commented Sep 25, 2022 at 21:17
  • $\begingroup$ I think I just figured out the answer: let $\sigma(T')$ and $\sigma(T)$ be the $\sigma$-algebra generated by $T',T$, then $\sigma(T)\subseteq \sigma(T')$, hence there is a Borel-measurable function $h$ such that $T=h\circ T'$. $\endgroup$ Commented Sep 25, 2022 at 21:27

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Let $h$ be non-measurable and let $g$ be constant. Then so is $f$. In particular both $f$ and $g$ are measurable but $h$ is not.

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