# 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.

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

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?

• No. See this for a counter example math.stackexchange.com/questions/283443/… . The point is that if we look at $\{f(B) : B \in \mathcal{B}\}$, then this is a sigma algebra containing the Borel sigma algebra $\mathcal{B}$. I'll let you think about why it is exactly the Borel sigma algebra. – mathworker21 Oct 6 '18 at 23:06
• I think I have this now. We show that {f(B)} is closed under complementation and countable intersections and contains all open sets. Then since B is the smallest such sigma algebra, we have that B is contained in {f(B)}. Then we use the fact that f is a homeomorphism and the property that f is Borel measurable iff the preimage of any Borel set under f is Borel to show that {f(B)} is contained in B. Then they are equal. Does this sound about right? – mrmingus Oct 6 '18 at 23:43
• Note above, I meant to say countable unions, although I think one implies the other anyway. – mrmingus Oct 6 '18 at 23:45
• how do you know $f$ is Borel measurable? what I was thinking is that since $f$ is bijective, you can just reverse the argument. $f^{-1}(f(B))$ contains $f(B)$, but we know $f^{-1}(f(B)) = B$, so we're done. – mathworker21 Oct 7 '18 at 0:22
• If f(y) = y+x, f is Borel measurable since f continuous. From the text I'm using, the set of Borel measurable functions on R is defined to be the smallest set of all real valued functions that contains all continuous functions and is closed under pointwise limits. So f continuous implies f Borel measurable. In any case, thank you for the response. – mrmingus Oct 7 '18 at 0:55