This question is related to another one 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-Borelanalytic 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
finitefor 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!

injectiveBorel function, hence its image $\pi(A) = \pi(M)$ is Borel. $\endgroup$ – Nate Eldredge Nov 2 '16 at 6:45