Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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
add comment

2 Answers 2

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

share|improve this answer
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.