Maybe this is just a stupid question, but I have tried very hard and get no luck. Here is the question:

Let $S_{k-1}$ be the unit sphere in $R^k$, i.e., the set of all $u\in R^k$ whose distance from the origin 0 is 1. So it is true that every $x\in R^k$, except for $x=0$, has a unique representation of the form $x=ru$, where $r$ is a positive real number and $u\in S_{k-1}$. Thus $R^k- \lbrace 0\rbrace$ may be regarded as the cartesian product $(0,+\infty)\times S_{k-1}$. So there is a one-to-one mapping $\phi$ of $R^k$ onto $(0,+\infty)\times S_{k-1}$.

Let $\mathfrak{B}(X)$ denote the $\sigma$-algebra generated by open sets in $X$. Let $\mathfrak{B}(X)\times \mathfrak{B}(Y)$ denote the $\sigma$-algebra generated by the form $A\times B$, which $A\in \mathfrak{B}(X)$ and $B\in \mathfrak{B}(Y)$.

My problem: if $A\in \mathfrak{B}(R^k-\lbrace 0 \rbrace)$, is it true that $\phi(A) \in \mathfrak{B}((0,+\infty))\times \mathfrak{B}(S_{k-1})$.

I think this problem is true, trivially. If this problem is just a well-known result, please show me some references. Thank you very much!

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    $\begingroup$ IIRC, you can find this in Folland's Real Analysis book. $\endgroup$ – Bill Johnson May 2 '11 at 5:14

The point here is that $\phi^{-1}:(0,+\infty)\times S_{k-1}\to\mathbb R^k\setminus\{0\}$ is actually continuous and thus Borel-measurable. Hence $\phi$ maps Borel sets to Borel sets.

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    $\begingroup$ Stefan's answer is correct but it might mislead someone to think that images of Borel sets under Borel mappings are always Borel. That's not true in general (not even for continuous mappings), but it is true when the mapping is one-to-one (or even just countable-to-one). That's a non-trivial result, but in the present situation, where $\phi$ is a homeomorphism, things are much easier. The notion of "Borel set" is defined in purely topological terms, so it is preserved by homeomorphisms. $\endgroup$ – Andreas Blass May 2 '11 at 8:20
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    $\begingroup$ Andreas, I was hoping that I had put enough emphasis on the fact that $\phi^{-1}$ is Borel measurable, and not just $\phi$. $\endgroup$ – Stefan Geschke May 2 '11 at 10:34
  • $\begingroup$ The comment is correct but it does not correctly answer the original question (it may answer the question the proposer wanted to answer!) The answer is that there is no reason that $\phi(A)$ should be the product of two measurable sets (just take a square of side 1 centred at (1,0) for example). $\endgroup$ – Anthony Quas May 2 '11 at 12:45
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    $\begingroup$ @Anthony: The question was not about products of Borel sets (which certainly are not preserved by $\phi$) but about the sigma-algebras they generate. The point is that, in products of nice spaces like these (second countable ones), the sigma-algebra generated by such products is also the sigma-algebra generated by the open sets, so it's a topological invariant. I guess your objection is that what I just said should have been in my original comment; I agree. $\endgroup$ – Andreas Blass May 2 '11 at 16:29

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