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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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What is $\gamma(x)$? And on which type of space are you working? –  Davide Giraudo Jun 28 '12 at 14:56
And which definition of Radon measure are you using? –  Davide Giraudo Jun 28 '12 at 15:06

1 Answer 1

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.

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