Let $F$ be a closed set in the plane $\mathbb{R}^2$, define $F_y=\{x\in\mathbb{R}|(x,y)\in F\}$, is $\bigcup_{y\in \mathbb{R}}F_y$ a borel set in $\mathbb{R}$? Intuitively, it is just like compressing a closed set in the plane into a set in the real line. I become interested in this problem for this is the last step that I need to finish an exercise, but I struggle to figure out whether it is true or not. Any help will be appreciated.
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$\begingroup$ The set $\bigcup_y F_y$ is in fact just the image of $F$ under the projection onto the first factor - AKA, "the projection of $F$". This lead me to the answer here (just the bit at the end), which is the same as the answer below - although perhaps a bit easier to motivate/easier to generalise/cleaner if you know facts like "$\Bbb R^2$ is $\sigma$-compact", and "the image of a compact set is compact", and "a compact set in $\Bbb R$ is closed". Often in measure theory you argue by "approximating" something complicated by some simpler things. $\endgroup$– Izaak van DongenMar 19 at 10:31
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It is actually an $F_{\sigma}$ set (i.e. a countbale union of closed sets). In fact, $\{x: x \in F_y \,\,\text {for some } y \in [-N,N]\}$ is a closed set for each $N$.
Indeed, if $(x_n,y_n) \in F$ with $|y_n|\leq N$ and $x_n \to x$ then there is a subsequence $y_{n_i}$ converging to some $y \in [-N,N]$ and $(x,y) \in F$.
