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Can you help me with the following?

For $A \subset \mathbb{R}^d$ and $\mathbb{a}> 0$, let $\mathbb{a}A:=\{\mathbb{a}x\mid x\in A\}$. If $A$ is a Borel set in $\mathbb{R}^d$, then $\mathbb{a}A$ is also a Borel set.

I thought about using the good set principle and I already showed that $aA$ is a $\sigma$-Algebra. The only thing missing, is to show that $aA$ contains the generator of the Borel-Algebra.

But we've never learned how to find it and I don't have an idea how to do that. Thanks for your help.

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  • $\begingroup$ Well if you showed $aA$ is Borel, then aren't you done? $\endgroup$ – user332239 Dec 22 '16 at 21:27
  • $\begingroup$ Sorry, I meant I did show that $aA$ is a $\sigma$-Algebra. Just edited it. $\endgroup$ – Tobi92sr Dec 22 '16 at 21:35
  • $\begingroup$ I changed $\mathbb{a}A$:={$\mathbb{a}x$|$x\in A$} to $\mathbb{a}A:=\{\mathbb{a}x \mid x\in A\}.$ One shouldn't alternate in and out of MathJax like that. Just keep all the math notation inside the MathJax tags. That way you don't get font mismatches and lack of proper spacing and incorrect alignment. $\endgroup$ – Michael Hardy Dec 22 '16 at 21:40
  • $\begingroup$ There are several generators for the Borel $\sigma$-algebra - I find it odd that you have not seen examples of some of them in lectures or your book, $\endgroup$ – msx Dec 22 '16 at 22:20
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    $\begingroup$ How did you show $aA$ is a $\sigma$-algebra? It is definitely not since, say, $\emptyset$ isn't an element. $aA$ is like a scaled version of your original set $A$. I think your whole approach is a bit off here. You want to show that $aA$ is a Borel set. $\endgroup$ – user332239 Dec 22 '16 at 22:44
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As mentioned above you should recap expressions like "Borel set", "$\sigma$-Algebra" and "generator".

If you did consider the function $f(x) = \frac{1}{\mathbb{a}}x$ which is continuous hence measurable and it holds $$\mathbb{a}A = f^{-1}(A)$$ and so $\mathbb{a}A$ is a Borel set because $A$ is.

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