If $B$ Borel measurable, $x \in \mathbb{R}$, then $B + \{x\}$ Borel measurable

Question

I imagine that this is pretty obvious, but I'm missing something. It's part of a larger proof to show that if $B$ Borel, $A$ countable, then $B+A$ Borel. If I can get $B + \{x\}$ Borel, the rest should follow fairly easily. Of course, $\{x\}$ is Borel, so I've tried writing $\chi_{(B+\{x\})}$ as various combinations $\chi_{B}$ and $\chi_{\{x\}}$, but to no avail. I've also tried working with set identities to write $B+\{x\}$ as the union/intersection/complement, etc of sets that are Borel measurable. A pointer in the right direction would be awesome.

Answer 1

Hint: for any $x \in \mathbb{R}$, the map $f: \mathbb{R} \to \mathbb{R}$ given by $f(y) = y+x$ is a homeomorphism.

Answer 2

First off, thanks for the answers. I appreciate it. I tried to work out the following proof based off of your suggestions. Can someone please tell me if this is correct, and if not, why?

Suppose $B$ is Borel, $x$ $\in$ $R$. Let $f(y)=y-x$. Then $\chi_{(B+\{x\})}$$(y)$ $=$ $\chi_{B}$$(f(y))$ $=$ $\chi_{B}$$(y-x)$, since $y$ $\in$ $B+\{x\}$ $\iff$ $y-x$ $\in$ $B$. Since $\chi_B$ Borel measurable and $f(y)$ continuous (and thus Borel measurable), $\chi_{B}$$(f(y))$ is Borel measurable.

At this point, I haven't proven that the composition of two Borel measurable functions produces a Borel measurable function. Perhaps that is where the notion of homeomorphism comes in... I have to look at it further. In any case, is this the right path?