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Suppose that $\pi : (X, \Sigma_X) \to (Y, \Sigma_Y)$ is a measurable surjection between two measurable spaces. Let $F = \pi^{-1} (\Sigma_Y)$ be the sub $\sigma$ algebra of $\Sigma_X$. Let $f : X \to (\mathbb{R}, B)$ be a measurable function to the reals with the Borel sigma algebra (or to any Hausdorff space with the Borel sigma algebra).

Question: Is $f$ $F$ measurable iff there is a measurable function $g : Y \to \mathbb{R}$ so that $f = g \circ \pi$?

I'm fairly certain I can prove this, and I can provide my argument if someone wants to see it. However, because these are all tricky measure theory technicalities, I'm asking here to be sure. Maybe someone knows a counter example off the top of their head.

Rmk: The surjectivity of $\pi$ is necessary, a counter example is the inclusion of a non-measurable set $X$ into $Y = \mathbb{R}$, with $f = 1_{X}$ and the induced sigma algebra on $X$. Edit: Nevermind, I was being silly. $1_Y$ also pulls back to $1_X$ -- no uniqueness in this case. :)

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    Could you elaborate on your remark re surjectivity of $\pi$? The claim you make seems to be theorem 4.41 in Aliprantis and Border, except that A&B do not require that $\pi$ be surjective. (They also do not assume a $\sigma$-algebra on $X$ other than what is already induced by $\pi$, but I don't see why that should matter.) – 2017-02-11
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    @TheoreticalEconomist Thanks for the reference. I was being silly about the counter example -- of course the indicator on $Y$ also also pulls back to $1_X$. – 2017-02-11
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    I'm probably a bit late, but this is called the [Doob-Dynkin lemma](https://en.wikipedia.org/wiki/Doob%E2%80%93Dynkin_lemma). The surjectivity of $\pi$ is not necessary. You can replace $(\mathbb{R}, B)$ with more general spaces, but "any Hausdorff space with the Borel sigma algebra" won't do. See [this MO question](http://mathoverflow.net/q/263863/90906) for details about the conditions. – 2017-03-18

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