How should I prove that $\pi +\mathbb{Q}$ is an element of the Borel $\sigma$-algebra?

Question

The definition of the Borel $\sigma$-algebra is the $\sigma$-algebra generated by the open sets.

How should we prove that $\pi +\mathbb{Q}$ is an element of the Borel $\sigma$-algebra based on the definition above? $\mathbb{Q}$ is the set of all rational numbers.

Answer 1

Note that for each $x \in \Bbb R$, we have $\{x\} = \bigcap_{n \geq 1} ]x-1/n, x+1/n[$, so that $\{x\}$ is Borel. Hence $\Bbb Q = \bigcup_{x \in \Bbb Q}\{x\}$ is Borel (countable union of Borel sets). Follows that $\pi + \Bbb Q$ is Borel, since translations of Borel sets are Borel sets (proof: $\mathcal{B}' \doteq \{ \pi + A \mid A \mbox{ is Borel} \}$ is a $\sigma$-algebra containing all Borel sets).

Answer 2

$\pi+\mathbb{Q}$ is a countable set. Since each singleton $\{\pi+r\}$ for $r$ rational is in the Borel $\sigma$-algebra (why?), and countable unions of elements of a $\sigma$-algebra are also in that $\sigma$-algebra, it follows that $\pi+\mathbb{Q}$ is in it too.