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.

Solution 1:

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.

Solution 2:

Let translation be $T_a(B)=a+B$. Then it is easy to show that $T_a(\mathcal{B}(\mathbb{R}))$ is a $\sigma$-field. It can also be easily shown that $T_a(\mathcal{B}(\mathbb{R}))$ contains the field of finite disjoint unions of right semi-closed intervals (say $\mathcal{F}_0$). and $\mathcal{B}(\mathbb{R}) = \sigma(\mathcal{F}_0)$.

Therefore, $$\mathcal{B}(\mathbb{R}) \subset T_a(\mathcal{B}(\mathbb{R})).$$

Now, to prove that $T_a(\mathcal{B}(\mathbb{R})$) $\subset$ $\mathcal{B}(\mathbb{R})$ note that $T_{-a}(T_a(\mathcal{B}(\mathbb{R}))=\mathcal{B}(\mathbb{R})$ $\forall a$.

Suppose that $ \exists \omega \in T_{a}(\mathcal{B}(\mathbb{R}))$ such that $\omega$ $\notin \mathcal{B}(\mathbb{R})$. But we have $T_{-a}(\omega)$ $\in \mathcal{B}(\mathbb{R})$. Since $a$ can be replaced by $-a$, we have our proof.