We consider the Euclidean space version of the Doob Dynkin lemma. In particular we first consider the real line, and the statement of the lemma becomes
If $X,Y:\Omega\to \Bbb R$ are random variables, then $Y$ is measurable with respect to $\sigma(X):=\{X^{-1}(B)\mid B\in\mathcal B(\Bbb R)\}$ iff there exists Borel-measurable $f:\Bbb R\to \Bbb R$ such that $Y=f(X)$.
Here's an outline for a proof in which $X$ is surjective, which hopefully may inspire some thoughts for the non-surjective case. The non-trivial direction is the existence of $f$. If we assume surjectivity of $X$ then the proof is easy. Let $Q_m^n:=[m/2^n,(m+1)/2^n)$ and $B_m^n\in\mathcal B(\Bbb R)$ such that $Y^{-1}(Q_m^n)=X^{-1}(B_m^n)$. By surjectivity we would have that for any fixed $n$, $\uplus_mB_m^n = \Bbb R$ and hence we can define the following function: $$f_n:\Bbb R\ni x\mapsto m/2^n\in\Bbb R,\quad x\in B_m^n.$$ And it is evident that $f_n$ is ever increasing at each $x\in\Bbb R$ and defining $f:=(\lim)f_n$ we can show $f(X)=Y$.
However, in the absence of the surjectivity of $X$, there seems to be no obvious way to "consistently" define the part of $B_m^n$ that is outside $X(\Omega)$ so that $\{B_m^{n+1}\}$ can be a refinement of $\{B_m^n\}$.
Is it possible to get around this difficulty? Also, I know there is a monotone class technique which can prove this result, but that's not what I want.
Cheers.
PS: it might be worth noting that when $X$ fails to be surjective, $f$ can't be uniquely determined (of course..)
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$\begingroup$If you mean the following by a "monotone class technique", I will delete my answer. Otherwise, I think the following proof is nice:
We know (see e.g. Measurable non-negative function is infinite linear combination of $\chi$-functions) that if $f \geq 0$ is $\sigma (X ) $ measurable, then $f = \sum_n c_n 1_{E_n} $ with $c_n \geq 0$ and $E_n \in \sigma (X) $. In particular, the series converges everywhere.
By definition of $\sigma (X) $, $E_n = X^{-1}(M_n) $ for suitable Borel sets $M_n $. Now, define $g_0 := \sum_n c_n 1_{M_n} $ and let $g := 1_{g_0 <\infty} \cdot g_0$ and note that $g $ is real valued and Borel measurable. It is not hard to see $f = g_0 \circ X $ and then $f = g\circ X $, since $g_0 (X (\omega)) =f (\omega)<\infty $ for all $\omega$.
For the general case, split $f = f_+ - f_-$ and apply the preceding proof.
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