Timeline for Question on application of Tonelli's theorem in convolution (topological groups)
Current License: CC BY-SA 3.0
10 events
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| Dec 15, 2016 at 17:43 | comment | added | Giuseppe Negro | By the way, the remark on Rudin's book is exactly the same as @NateEldredge 's. There are some more details on Folland's "Real Analysis", second edition, pag. 240 (and Exercise 5, pag.245). This exercise, in particular, directly shows that $(x, y)\mapsto f(x-y)g(y)$ is Lebesgue measurable whenever $f, g$ are, without passing from the Borel sigma algebra, so it might apply to the general case of $G$ not being second countable also. (HTH) | |
| Dec 8, 2016 at 17:01 | comment | added | PhoemueX | @NateEldrege: I am not sure that the continuity suffices. For this you would need to know that the product sigma algebra on $G \times G $ is identical to the Borel sigma algebra on $G \times G $. This is true if $G $ is 2nd countable, but can fail in general IIRC. | |
| Dec 8, 2016 at 15:34 | comment | added | Nate Eldredge | This is a standard fact from measure theory: if $\mathcal{B}$ is a $\sigma$-algebra and $\mathcal{B}'$ is its completion with respect to some measure $\mu$, then for any real-valued $\mathcal{B}'$-measurable function $f$, there is a $\mathcal{B}$-measurable function $f_0$ with $f=f_0$ $\mu$-almost everywhere. In particular they would represent the same element of $L^1$. | |
| Dec 8, 2016 at 7:39 | comment | added | TheGeekGreek | @NateEldredge Thanks a lot. Just to be sure: why exactly can I make the assumption that $f$ and $g$ are Borel? I do not quite see the w.l.o.g. | |
| Dec 7, 2016 at 23:27 | comment | added | Nate Eldredge | Without loss of generality, choose versions of $f,g$ which are Borel. The map $(x,y) \mapsto y^{-1}x$ is jointly continuous, since we are in a topological group, hence jointly measurable. Hence so is $(x,y) \mapsto g(y^{-1}x)$. And the product of (jointly) measurable functions is (jointly) measurable. | |
| Dec 7, 2016 at 22:57 | comment | added | Giuseppe Negro | It must be there nearby. I forgot Rudin treats convolution on $R$ and relegates the case of $R^n$ to the exercises. | |
| Dec 7, 2016 at 22:52 | comment | added | TheGeekGreek | Thanks for your remark. Where do I find this exactly? I have the third edition and on page 170 he defines convolution on $\mathbb{R}$. Where is the remark for $\mathbb{R}^n$? | |
| Dec 7, 2016 at 22:31 | comment | added | Giuseppe Negro | There is a remark on this on Rudin's "Real and complex analysis", when he speaks of convolution on $R^n$. I bet the same argument works in the general case. | |
| Dec 7, 2016 at 22:06 | history | edited | TheGeekGreek | CC BY-SA 3.0 |
added 118 characters in body
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| Dec 7, 2016 at 21:49 | history | asked | TheGeekGreek | CC BY-SA 3.0 |