Let $(\Omega,\mathcal A,\operatorname P)$ be a probability space and $X,Y$ be real-valued random variables on $(\Omega,\mathcal A,\operatorname P)$.
If $X$ and $Y$ are dependent, can we always find a Borel measurable $f:\mathbb R\to\mathbb R$ with $Y=f(X)$?
This is actually equivalent to the claim that $Y$ is $\sigma(X)$-measurable.
I guess the answer in general is "no", if we understand "dependent" as "not independent" and I guess $Y=f(X)$ for some Borel measurable $f:\mathbb R\to\mathbb R$ is a particular kind of dependence ... But what could dependence mean otherwise?
Let $A,B,C$ be a partition of the sample space where each of the three sets has positive probability. Let $X=1$ on $A \cup B$ and $0$ on $C$. Let $Y=2$ on $A$, $3$ on $B$ and $1$ on $C$. Note that $P(X=1,Y=1) =0$ but $P(X=1) P(Y=1) >0$. Hence $X$ and $Y$ are not independent. However there is no $f$ such that $Y=f(X)$; this is because if such a function exists then $X=1$ implies $Y=f(1)$. But $Y$ take two distinct values on the set $X=1$.