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?

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    $\begingroup$ 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. $\endgroup$ – mathworker21 Oct 6 '18 at 23:06
  • $\begingroup$ 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? $\endgroup$ – mrmingus Oct 6 '18 at 23:43
  • $\begingroup$ Note above, I meant to say countable unions, although I think one implies the other anyway. $\endgroup$ – mrmingus Oct 6 '18 at 23:45
  • $\begingroup$ 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. $\endgroup$ – mathworker21 Oct 7 '18 at 0:22
  • $\begingroup$ 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. $\endgroup$ – mrmingus Oct 7 '18 at 0:55

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