For $X$ and $Y$ defined on some probability space $(\Omega,\mathcal{F},\mathbf{P})$, $Y$ is $\sigma(X)$-measurable iff there is a Borel function $f$ such that $Y=f(X)$.
My question is, given $X$ and $Y$ and knowing that $Y$ being $\sigma(X)$-measurable, is the Borel function $f$ unique? Can there be more than one such Borel functions fulfill the condition?
It helps to rephrase this question in terms of pure measure theory: let measurable $X,Y\in\Omega\to\Psi$ be such that $Y=f\circ X$ for some Borel $f$. Is $f$ unique?
This immediately makes it clear that Borelness is not relevant to this problem. The answer is the well-known theory of lifts in the category of sets.
If $X$ is not surjective, the answer is clearly no; we can make arbitrary changes to $f$ on the complement of $\text{im}{(X)}$.
If $X$ is surjective, then $f$ is unique by linearity. For, suppose $Y=f\circ X=g\circ X$; then, by subtracting, we have $(f-g)\circ X=0$. As $X$ is surjective, $f-g=0$.