Uniqueness of mesurable function composition

Question

Let $h$ be a $\left( \Omega\mathcal{,F} \right) \rightarrow \left( S,\mathcal{E} \right)$ measurable function, and $g$ be a $\left( \Omega\mathcal{,F} \right) \rightarrow \left( \mathbb{R,}\mathcal{B}\left( \mathbb{R} \right) \right)$ measurable function, where $\mathcal{B}\left( \mathbb{R} \right)$ is the Borel $\sigma$-algebra of $\mathbb{R}$. How do we prove there exists an almost surely unique $\left( \mathbb{R,}\mathcal{B}\left( \mathbb{R} \right) \right) \rightarrow \left( \mathbb{R,}\mathcal{B}\left( \mathbb{R} \right) \right)$ measurable function $f$ s.t. $f \circ g = h$?

Answer 1

This is horribly false. For instance, if $g$ is constant, then $f\circ g$ is constant for any $f$, so such an $f$ cannot exist unless $h$ is constant. Moreover, even when $h$ is constant, $f$ will not be unique (up to sets of measure $0$) since only the value of $f$ at the constant value of $g$ matters.