I'm reading a proof that all finite equivalence relations are smooth, which goes as follows:
Let $E$ be a finite Borel equivalence relation on a Polish space, we want to find a Borel function $f\colon X\to X$ reducing $E$ to $\mathrm{Id}_X$. Let $<$ be a Borel total order on $X$ and define $f(x)$ to be the $<$-least element of $[x]_E$.
Clearly $xEy\iff f(x)=f(y)$, but it's not clear to me why $f$ is Borel. I'm trying to argue that $f^{-1}([x,\infty))$ is Borel for all $x\in X$, but I can only show that is $\mathbf{\Pi}^1_1$ and I'm not seeing why it must be $\mathbf{\Sigma}^1_1$ as well, hence my question: why is this function $f$ Borel?
From the comments:
Use Feldman-Moore to fix a countable $G$ Borel acting on $X$ such that $E={\sim_G}$. Now $f(x)=y$ iff $(x,y)\in E$ and $\forall g\in G\,(gy\ge y)$, so $f$ is Borel.