Slice of a Borel set in $\mathbb{R}^2$ is Borel

Question

Assume that $E$ is Borel subset of $\mathbb{R}^2$. Show that for every $y \in \mathbb{R}$ , the slice $E^y = \{ x \in \mathbb{R}|(x, y) \in E \}$ is a Borel subset of $\mathbb{R}$.

I would be showing this by considering the family $F$ of all sets $E \subset \mathbb{R}^2$ that have the property of $E^y$ is borel. That is Let $F = \{ E \subset \mathbb{R}^2|E^y$ is a Borel set in $\mathbb{R} , \forall y ∈ \mathbb{R} \}$. namely that all slices of E are Borel sets. Then showing that $F$ is sigma algebra , i.e. contains all open subsets of $\mathbb{R}^2$ and that $F$ is closed under countable unions, countable intersections, and complements.

Then showing that $F$ contains open sets.

Is the above logic correct?

I can show that $F$ is sigma algebra but I have problem to show that it contains open sets. How do I show that?

Answer 1

Yes, your logic is correct and a rather interesting way to go about it. The Borel sigma algebra is often defined by the property that any sigma algebra containing the open sets contains every Borel set, which is the property that you are using.

It happens that if $U$ is an open subset of $\mathbb R^2$ then, for every $y\in \mathbb R$, it holds that the set $U^y=\{x\in \mathbb R : (x,y) \in U\}$ is open - hence Borel - implying that $U\in F$. You can see this in two ways: Firstly, directly by definition, note that if $x\in U^y$ then there is an open ball around $(x,y)$ contained in $U$, but the slice of that open ball is an open interval around $x$, which directly shows that every point in $U^y$ has a neighborhood around it. The second way, as in @Reveillark's answer, is to define the function $f_y(x)=(x,y)$ and to note that $U^y = f_y^{-1}(U)$. Since $f_y$ is continuous, the preimage of an open set is open.

Note, more generally, one can argue that, if $f$ is any function between topological spaces such that $f^{-1}(U)$ is Borel whenever $U$ is open, then $f^{-1}(U)$ is Borel whenever $U$ is Borel. This statement is a nice thing to prove and can probably be gotten by simplifying the work you've already done (which amounts to some statement about a family of such functions).

Answer 2

Here’s an easier argument:

Fix $y$, and consider the function $f(x)=(x,y)$. This function is Borel (it’s obviously much more), so $f^{-1}(E)$ is Borel when $E$ is Borel. But $f^{-1}(E)=E^y$.