Q:

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

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.

A:

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

