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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.

mwt
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mrmingus
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2 Answers2

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

mathworker21
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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?

mrmingus
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    No. See this for a counter example https://math.stackexchange.com/questions/283443/is-composition-of-measurable-functions-measurable . 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 06 '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 06 '18 at 23:43
  • Note above, I meant to say countable unions, although I think one implies the other anyway. – mrmingus Oct 06 '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 07 '18 at 00:22
  • 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 07 '18 at 00:55
  • Actually, I think that your argument to show reverse inclusion generalizes better to other cases. I worked out the proof for B+{x} and then extended that to B+A being Borel given B Borel, A countable. Now I am looking at the case where B Borel, A open, to show that B+A Borel. My approach is to show that B+(c,d) Borel, and then use the fact that any open set can be expressed as the countable union of disjoint open intervals to show that B+A is Borel. Do you think the approach of defining f(y) = (y+c,y+d) and following an analogous path is worth pursuing? – mrmingus Oct 07 '18 at 11:07
  • I mean, I'm going to pursue it any way, at least for a while... but am I barking up the wrong tree? – mrmingus Oct 07 '18 at 11:08