# 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.

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This amounts to showing that the family of invariant Borel sets forms a $\sigma$-algebra that contains all Borel sets. –  Michael Greinecker Oct 10 '12 at 5:51

Translation ($T_a(x) = x+a$) is continuous, hence Borel measurable. Hence $B+a = T_{-a}^{-1} B$ is (Borel) measurable.

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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.

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