This question is related to [another one](https://mathoverflow.net/questions/253478/can-the-integration-of-integrable-sections-of-a-measurable-function-of-two-varia) that I asked two days ago.



> 
> **Question.** Does there exist a Borel subset $ M $ of $ \mathbb{R}^{2} $ with
>  the following two properties?
> 
> 
> * The projection of $ M $ onto the first component is a **non-Borel** analytic
>  subset of $ \mathbb{R} $.
> * Every vertical cross-section of $ M $ is finite, i.e., the set
>  $ \{ y \in \mathbb{R} \mid (x,y) \in M \} $ is **finite** for every
>  $ x \in \mathbb{R} $.
> 
> 
> 


An affirmative answer to this question will provide a counterexample in the topic of measure theory, as explained in the linked post. It therefore holds some importance.


Thank you very much for your help!



Is there a Borel subset of $ \mathbb{R}^{2} $, with finite vertical cross-sections, whose projection onto the first component is non-Borel?
No, no such set exists. This is a special case of the **Lusin–Novikov theorem**; see e.g. Kechris, *Classical Descriptive Set Theory*, Theorem 18.10.


In general, let $X,Y$ be standard Borel spaces, and suppose $M \subset X \times Y$ is Borel. (Here we are taking $X=Y=\mathbb{R}$.) For $x \in X$, let $M\_x = \{y : (x,y) \in M\}$ be the section of $M$ at $x$. The Lusin–Novikov theorem asserts that if all but at most countably many of the $M\_x$ are at most countable (i.e. $|\{x : |M\_x| > \aleph\_0\}| \le \aleph\_0$) , then $M$ has a Borel uniformization: there is a Borel set $M^\* \subset M$ such that $M^\*$ intersects every nonempty section $M\_x$ in exactly one point. In particular, the projection map $\pi : X \times Y \to X$, which is Borel, is injective when restricted to $M^\*$; so as a consequence (see Kechris Corollary 15.2) we have that $\pi(M^\*) = \pi(M)$ is Borel in $X$.


Any set $M$ satisfying your second condition certainly satisfies the hypothesis of Lusin–Novikov (since $\{x : |M\_x| > \aleph\_0\} = \emptyset$), so its projection is Borel, and thus it does not satisfy your first condition.


