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I have a question about Radon meaures: Given a Radon measures $ \mu_{1}, \mu_{2}$, both have compact support:

How to show that $\int \hat{\mu_{1}}(x)\,d\mu_{2}(x)=\int \hat{\mu_{2}}(x)\,d\mu_{1}(x)$ , where the "hat" means Fourier transform.

Thanks for help in advance.

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    An$d$ which $d$efinition of Radon measure are you using?2012-06-28

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It's an application of Fubini's theorem (the Fourier transform are well-defined since the measures have compact support). We have $\int \widehat{\mu_1}(x)d\mu_2(x)=\int\int e^{iyx}d\mu_1(y)d\mu_2(x).$ Since $\mu_1(\mathbb R)\times \mu_2(\mathbb R)$ is finite (it's in fact the product of the measure of the respective supports, which is finite by definition of Radon measure), we can switch the two integrals to get the wanted formula.