Equivalent of Lusin's Theorem in Borel setting Let $X$ be a Polish space, $\mathcal B$ the sigma-algebra 
of Borel sets.  Let  $E$  be an 
aperiodic countable Borel equivalence relation on 
$X \times X$ (this means that every class of equivalence
is countably infinite). A set $C\in \mathcal B$ is called a complete
section for $E$, if $\forall x \in X$ $\exists y \in C$ such that 
$(x, y) \in E$. 
$\textit{Question:}$ Let $f : X \to \mathbb R$ be a Borel real-valued function.
Is there a complete section $C$ such the restriction of $f$ on the set
$C$, $f|_C$, is a continuous function.
 A: The answer is No. 
A suitable counterexample can be constructed as follows.
On the real line $\mathbb R$ consider the equivalence relation $E=\{(x,y)\in\mathbb R\times \mathbb R:x-y\in\mathbb Q\}$. 
Fix a countable base $\{U_n\}_{n\in\omega}$ of the topology on the real line. Take a countable set $X=\{x_n\}_{n\in\omega}\subset\mathbb R$ such that $x_n-x_m\notin \mathbb Q$ for any distinct $n,m\in\omega$. 
In the real line $\mathbb R$ consider the $G_\delta$-set 
$$G:=\mathbb R\setminus\bigcup_{n\in\omega}(\mathbb R\setminus U_n)\cap(x_n+\mathbb Q)=(\mathbb R\setminus(X+\mathbb Q))\cup\bigcup_{n\in\omega}U_n\cap(x_n+\mathbb Q)$$ and the Borel function $f:G\to\mathbb R$
$$f(x)=\begin{cases}
n&\mbox{if $x\in U_n\cap (x_n+\mathbb Q)$ for some $n\in\omega$};\\
0&\mbox{otherwise}.
\end{cases}
$$
It is easy to see that the equivalence relation $E_G:=E\cap(G\times G)$ on $G$ has
no complete section $C$ with continuous restriction $f|C$; moreover, for any complete section $C$ of $E_G$, the restriction $f|C$ has no continuity points.

There are (a bit more involved) counterexamples even for very good equivalence relations.
To construct a suitable example, take the convergent sequence $S:=\{0\}\cup\{2^{-n}:n\ge0\}\subset\mathbb R$ and on the compact zero-dimensional space $X:=S^\omega\times S$ consider the equivalence relation 
$$E=\{((x,y),(x',y'))\in X\times X:x=x'\}.$$ Observe that the quotient map $q:X\to X/E=S^\omega$ is open and closed.
Now take any bijective map $p:S\to D$ to a countable space $D$. Its countable power $P:S^\omega\to D^\omega$, $P:(x_n)_{n\in\omega}\mapsto (p(x_n))_{n\in\omega}$, is known as the Pawlikowski function and is a standard example of a Borel function of the first Baire class, which is not $\sigma$-continuous.
More precisely, the Pawlikowski function has the property that a subset $C\subset S^\omega$ is nowhere dense in $S^\omega$ if the restriction $P|C$ is continuous.
We claim that the Borel function $f:=P\circ q:X\to D^\omega$ and the equivalence relation $E$ yield a counterexample to the question posed by Shrey. Indeed, assume that the equivalence relation $E$ has a complete section $C$ with continuous restriction $f|C$. For every $s\in S$ let $C_s:=\{x\in S^\omega:(x,s)\in C\}$ and observe that the continuity of the restriction $f|C$ implies the continuity of the map $P|C_s$ and hence nowhere density of $C_s$ in $S^\omega$. Since the space $S^\omega$ is compact and hence Baire, we conclude that  $\bigcup_{s\in S}C_s\ne S^\omega$,  which contradicts the choice of $C$ as a complete section of $E$.
