2
$\begingroup$

Let $G$ be a locally compact group which is a countable union of compact subsets and $\lambda$ a left Haar measure on $G$ (i.e. $(G,\lambda)$ is $\sigma$-finite). Let $f,g \in L^1(G)$. Then I have a question on the following application of Tonelli/Fubini: $$\int_G \int_G |f(y)||g(y^{-1}x)|d\lambda(y)d\lambda(x)=\int_G \int_G |f(y)||g(y^{-1}x)|d\lambda(x)d\lambda(y)$$ Why is this legitimate? I do not quite see why $|f(y)||g(y^{-1}x)|$ is measurable which actually reduces to the question why the function of two variables $f(y)g(y^{-1}x)$ is measurable.

Edit. This can be found in the third edition of the book classical fourier analysis by Loukas Grafakos, p. 20.

Improve this question
$\endgroup$
8
  • $\begingroup$ 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. $\endgroup$
    – Giuseppe Negro
    Commented Dec 7, 2016 at 22:31
  • $\begingroup$ 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 $\mathbb{R}^n$? $\endgroup$
    – TheGeekGreek
    Commented Dec 7, 2016 at 22:52
  • $\begingroup$ It must be there nearby. I forgot Rudin treats convolution on $R$ and relegates the case of $R^n$ to the exercises. $\endgroup$
    – Giuseppe Negro
    Commented Dec 7, 2016 at 22:57
  • $\begingroup$ 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. $\endgroup$
    – Nate Eldredge
    Commented Dec 7, 2016 at 23:27
  • $\begingroup$ @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. $\endgroup$
    – TheGeekGreek
    Commented Dec 8, 2016 at 7:39

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .