if $B \in \mathscr{B}(\mathbb{R}^n)$ then $x + B \in \mathscr{B}(\mathbb{R}^n) \; \forall x \in \mathbb{R}^n$

Question

I am trying to show that if $B \in \mathscr{B}(\mathbb{R}^n)$ then $x + B \in \mathscr{B}(\mathbb{R}^n) \; \forall x \in \mathbb{R}^n$.

To show this, it suffices to show that the set $$\mathscr{A}_x := \{B\in \mathscr{B}(\mathbb{R}^n): x + B \in \mathscr{B}(\mathbb{R}^n)\} \subset \mathscr{B}(\mathbb{R}^n).$$ I need to show first that $\mathscr{A}_x$ is a $\sigma-$algebra. How can I show that if $B \in \mathscr{A}_x$ then $B^c \in \mathscr{A}_x$, i.e. $x+A^c \in \mathscr{B}(\mathbb{R}^n)$?

Answer 1

If $B\in\mathscr A_x$ then $B=x+A$ for some $A\in\mathscr B(\mathbb R^n)$.

Then: $$-x+B^{\complement}=(-x+B)^{\complement}=A^{\complement}\in\mathscr B(\mathbb R^n)$$

(Prove the first equality by showing that both statements $y\in-x+B^{\complement}$ and $y\in(-x+B)^{\complement}$ are equivalent with the statement $y+x\notin B$)

So that: $$B^{\complement}=x+(-x+B^{\complement})\in\mathscr A_x$$

Answer 2

This is immediate from the fact that $x+B^c=(x+B)^c$ (if that equation is not obvious to you, try and prove it!).

Answer 3

The mapping $T_x(y) = x+y$ is a homeomorphism, hence $\{x\}+B =T_{-x}^{-1}(B)$ is Borel.

Answer 4

Let $\mathscr{T}:=\mathscr{T}({\mathbb R}^n)$ be the topology on ${\mathbb R}^n$. It is obvious that $\mathscr{T}$ is translation invariant, i.e. $\Omega\in \mathscr{T}$ implies $\Omega+ x\in\mathscr{T}$ for all $x\in{\mathbb R}^n$. Since $\mathscr{B}:=\mathscr{B}({\mathbb R}^n)$ is constructed from $\mathscr{T}$ in a pure set theoretical way it immediately follows that $\mathscr{B}$ is translation invariant as well. You don't have to go through the motions.