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.



Equivalent of Lusin's Theorem in Borel setting
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](https://arxiv.org/pdf/0810.1391.pdf)* 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 $E$.


