Proving a set is Borel

Question

Prove that if $B \subseteq \mathbb R$ is a Borel set, then $B \times \mathbb R$ is a Borel in $\mathbb R^2$, i.e. $B \times \mathbb R \in B_2 $

I'm still trying to understand the definition properly. Please can someone show this. It would help me so much.

Thanks in advance!

Answer 1

A set is a Borel set if it is in the Borel $\sigma$-algebra.

So, $B \subseteq \Bbb R$ is a Borel set if $B \in \mathcal{B}(\Bbb R)$.

What is $\mathcal{B}(\Bbb R)$? It is the smallest $\sigma$-algebra containing all open subsets of $\Bbb R$ (i.e., the $\sigma$-algebra generated by the open sets).

You need to show $B \times \Bbb R$ is a Borel set in $\Bbb R^{2}$, i.e., $B \times \Bbb R \in \mathcal{B}(\Bbb R^{2})$.

But $\mathcal{B}(\Bbb R^{2})$ is the $\sigma$-algebra generated by the open subsets of $\Bbb R^{2}$. If you know that $\mathcal{B}(\Bbb R^{2}) = \mathcal{B}(\Bbb R) \times \mathcal{B}(\Bbb R)$ (where the RHS is the $\sigma$-algebra generated by measurable rectangles), then the problem becomes easy, because $B \times \Bbb R \in \mathcal{B}(\Bbb R) \times \mathcal{B}(\Bbb R)$.

Answer 2

The following proof is not the shortest, but (for me at least) makes things nicely intuitive.

We begin with a description of the Borel sets as a hierarchy:

• The $\Sigma^0_1$-sets are the open sets.

• For $\alpha$ a countable ordinal $>1$, the $\Sigma^0_\alpha$-sets are the sets of the form $B=\bigcup_{i\in\mathbb{N}} A_i$, where each $A_i$ is the complement of a $\Sigma^0_\beta$-set for some $\beta<\alpha$.

• Then we can prove:

  The Borel sets are the sets which are $\Sigma^0_\alpha$ for some countable ordinal $\alpha$.

(This is a good exercise if you haven't seen it before.)

Countable ordinals are often annoyingly abstract, but the hierarchy they provide allows us to use transfinite induction to prove results about Borel sets. For the specific question you're interested in, try proving the following stronger result by induction on $\alpha$:

Show that if $A$ is a $\Sigma^0_\alpha$-subset of $\mathbb{R}$, then $A\times\mathbb{R}$ is a $\Sigma^0_\alpha$-subset of $\mathbb{R}^2$.

Base case: if $A$ is open, is $A\times\mathbb{R}$ open?

Induction step: Suppose (IH) that $\alpha$ is some countable ordinal such that for every $\beta<\alpha$, if $A$ is a $\Sigma^0_\beta$ subset of $\mathbb{R}$ then $A\times\mathbb{R}$ is a $\Sigma^0_\beta$ subset of $\mathbb{R}^2 $. Now let $A$ be $\Sigma^0_\alpha$. By definition, this means $$A=\bigcup_{i\in\mathbb{N}} B_i,$$ where $B_i$ is the complement of a $\Sigma^0_{\beta_i}$ set $C_i$, $\beta_i<\alpha$. Look at $$A\times\mathbb{R}=[\bigcup B_i]\times\mathbb{R}=\bigcup[B_i\times\mathbb{R}]=\bigcup [C_i\times \mathbb{R}]^c.$$ Do you see how to continue?