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.
 A: 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.
A: Here, my aim is to show that $B + x = B$. My answer uses a slightly more simple notation than the other answers so hopefully it is some use! It is essentially the proof found here.
http://www.math.fsu.edu/~roberlin/maa5616.f13/translationinv.pdf
Firstly, we denote the collection of sets $M= \{A +x : A \in \Sigma\}$, the proof that this is a sigma-algebra can be found here Prove that if $A$ is a Borel set, then so is $x + A$, for every $x \in \mathbb{R}$. We let $\mathcal{M}$ denote the set of all $\sigma$-algebras.
Then $B_{\mathbb{R}} = \cap_{M \in \mathcal{M}}M$. 
Then $B_{\mathbb{R}} = \cap_{M \in \mathcal{M}}(M+ x) = \cap_{M \in (\mathcal{M}+x)} M $. It will suffice to show that $\mathcal{M} + x = \mathcal{M}$. We have that $M \in \mathcal{M} +x$ is a $\sigma$-algebra. Thus, $\mathcal{M}+x \subset \mathcal{M}$ and $\mathcal{M} = (\mathcal{M} + x) - x \subset \mathcal{M} + x$ and so $\mathcal{M} + x = \mathcal{M}$.
A: 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.
A: 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.
A: In this question I tried to prove that the set $B+t$ is an open set, so it is a member in the smallest sigma algebra that contains $B$, so $B+t$ will be a Borel set. Here I assume that $B = (a,b)$ and I must study when $t = a$ or $t = b$ and when $a<t<b$. Is this answer enough or not?
