Suppose $G$ is a locally compact abelian Hausdorff group (LCA), and $\lambda$ is the Haar measure on it. We all know the convolution of two $L^1(\lambda)$ functions $f$ and $g$ on $G$ is defined as $$f\ast g(x) = \int_G f(x-y) g(y) d\lambda(y).$$

In order to show that this is well defined (i.e. the integral converges) for almost all $x$, one utilizes Fubini's theorem to calculate $\int_G f\ast g(x) d\lambda(x)$ by reversing the order of integration. In order to justify the use of Fubini's theorem, many references in the literature do verify the hypothesis of $\sigma$-finiteness of the relevant measure (e.g. by restricting to the support), but I find no reference dealing with the question of measurability.

For me, the real question is, why the function $F(x,y) = f(x-y) g(y)$ on $G\times G$ is measurable with respect to the product measure $\lambda\times\lambda$, or even with respect to $\lambda\otimes\lambda$, the completion of it. For if we use $\mathfrak{B}$ to denote the $\sigma$-algebra of Borel sets, generally $\mathfrak{B}(G)\times\mathfrak{B}(G)$ is not the same as $\mathfrak{B}(G\times G)$ (the former can well be strictly smaller than the latter), nor is the completion of the former. So the trick by composing $F$ on the right by the homeomorphism $G\times G\to G\times G: (x,y)\mapsto (x-y,y)$ does not suffice as claimed in many books and articles.

We can of course consider $L^1(G)$ functions as complex measure on $G$, and consider the convolution of complex measures instead, but now the problem becomes a valid Radon-Nikodym theorem is missing in order to translate the complex measure back to functions, and the use of Fubini's theorem is still hard to avoid as one easily sees if he or she attempts to translate it back.

Of course if we further assume that $G$ is second countable or any open set in $G$ with finite Haar measure is $\sigma$-compact (e.g $(\mathbb{R}^n,+,0)$), the problem can be easily attacked. So we are in fact dealing with some kind of "pathology" here. But anyway, as maybe a stupid question as it is, it just keeps annoying me. So any help would be appreciated.

  • 2
    $\begingroup$ The default source when wants to be very careful about such issues is Volume 1 of Hewitt + Ross's Abstract Harmonic Analysis - I don't have a copy at hand right now, but if you have access to a copy, does it address your concerns? $\endgroup$ – Yemon Choi Jun 5 '14 at 15:13
  • $\begingroup$ @YemonChoi Yes, it helps a lot. Thanks. And for those who concerns this problem, it's section 20 of this book(cf. the above comment). $\endgroup$ – Hua Wang Jun 5 '14 at 17:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.