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Background:

Let $G$ be a locally compact group, and define $M(G)$ to be the vector space of all regular complex measures defined on $G$, normed by total variation.

For $\mu,\nu\in M(G)$, define the product $\mu * \nu$ as follows:

Define $\varphi\in C_{0}(X)^{*}$ by $$\varphi(f) = \int_{G}\int_{G}f(xy)d\mu(x)d\nu(y)\text{, for all }f\in C_{0}(X)$$

By the Riesz-Representation Theorem, there is a unique element (call it $\mu*\nu$) of $M(G)$ such that $\varphi(f) = \int_{G}fd(\mu *\nu)$.

This turns $M(G)$ into an algebra.


Proposition:

I'm reading a book which says that it is an immediate consequence of Fubini's Theorem that

$$\int_{G}\int_{G}f(xy)d\mu(x)d\nu(y) = \int_{G}\int_{G}f(xy)d\nu(y)d\mu(x)$$


My Question:

In order to apply Fubini's Theorem, we need $$\int_{G\times G}|f(xy)|d(\mu\times\nu) < \infty$$

which doesn't seem to follow from $f\in C_{0}(X)$.


Counterexample:

Take $G = \mathbb{Z}$, $f\in C_{0}(G)$ defined by $f(n) = \frac{1}{|n|}$. Then $\int_{G}f(x)d\mu(x) = \sum_{n=-\infty}^{\infty}\frac{1}{|n|} = \infty$.

From here it follows that for fixed $y\in G$, we have $\int_{G}f(xy)d\mu(x) = \infty$ so obviously the hypothesis of Fubini's Theorem is not satisfied.

But in this case, $f$ is non-negative, so the desired integral equation instead follows from Tonelli's Theorem, so all is not lost.


So does the Proposition follow from a different argument? or is an additional hypothesis on $G$ required?

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Thanks Zev, I didn't know it was that easy to type in bold! :) –  Kyle Schlitt May 26 '13 at 16:16
    
No problem - see here for other Markdown formatting tips –  Zev Chonoles May 26 '13 at 16:16

1 Answer 1

up vote 1 down vote accepted

I don't know if this is the answer, but it just occured to me that for $\mu\in M(G)$ to have finite norm, it must have finite total variation, which I think would be sufficient to guarantee the hypothesis of Fubini's Theorem. Maybe someone with more knowledge than I can confirm this?

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1  
Yes, that is the point. The convolution of two measures of finite variation is again a measure of finite variation. Convolution may also be defined for certain other measures as well, but that is not involved in the claim in the question. –  GEdgar May 26 '13 at 16:47
    
Thank you for confirming. Communicating a question to others seems to be quite illuminating. –  Kyle Schlitt May 26 '13 at 16:53

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