Prove Borel sigma-algebra translation invariant Can anyone explain: Let $B$ be a Borel set and $B + a = \{ x + a : x \in B\}$. Why is $B + a$ a Borel set?

I think I have to use some good set principle but not sure how to complete the proof.
Translation ($T_a(x) = x+a$) is continuous, hence Borel measurable. Hence $B+a = T_{-a}^{-1} B$ is (Borel) measurable.
Alternative proof:
Let ${\cal B}$ be the Borel sets, and let ${\cal B'} = \{ A \in {\cal B} | A+ \{x\} \in {\cal B} \}$. Note that ${\cal B'} \subset {\cal B}$ and a little bit of work shows that ${\cal B'}$ is a $\sigma$ field.
Since the open intervals are in ${\cal B'}$ we have ${\cal B} \subset {\cal B'}$ and hence they are equal.
Borel measurable functions mean the preimage of Borel set is Borel. Therefore we use subtraction instead of addition. 
    Take $B\subset \mathbb R$, and $t\in\mathbb R$, then take $f(b) = b-t$ $\forall b\in B$. This is continuous, thus Borel measurable. Therefore, $f^{-1}(B)$ is Borel measurable. However, $f^{-1}(B)$ is exactly $t+B$, therefore, $t+B$ is Borel measurable.