If $B\times \{0\}$ is a Borel set in the plane, then $B$ is a Borel set in $\mathbb{R}$. I'm trying to figure out how to prove the following "obvious" fact:

Let $B\times \{0\}\subset \mathbb{R}^2$ be a Borel set, then $B\subset \mathbb{R}$ is a Borel set.

The problem here is that I can't see a nice constructive way of expressing a Borel set, so I just can't intersect stuff with $\mathbb{R}\times\{0\}$ in some countable union/intersection, which is what I tried by starting by taking the countable family sets $[a,b]\times [c,d]$, $a,b,c,d\in \mathbb{Q}$ as the generators of the Borel sets of the plane.
Is there a nice technique of approaching results like this?

The map $f\colon (x,y)\mapsto \chi_{B\times\{0\}}(x,y)=\chi_B(x)\chi_{\{0\}}(y)$ is Borel measurable. As the map $\phi\colon (x,y)\mapsto (x,0)$ is Borel-measurable, so is $f\circ \phi$, which is $\chi_{B\times\Bbb R}$, hence $\chi_B$ is a Borel measurable function. 
