Inverse of Borel set is measurable?

Question

Just a short question.

I was reading this thread and got a question about one of the comments. In the second answer, @geekazoid said that

Also, $h−g$ is measurable, and $\{0\}$ is a Borel set, so $(h−g)^{−1}(0)$ is measurable.

The question is that why he mentioned about $\{0\}$ is a Borel set.

(Since I don't have enough reputation, I cannot but to make another question.)

Answer 1

Since $B:=\{0\}$ is Borel and $h-g$ is measurable, $(h-g)^{-1}(B)$ is Borel - measurable.

Use the Definitions of "Borel set" and "measurable function" !