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I am trying to prove the following seemingly obvious fact:

Let $\mu$ be a finite signed measure on $\mathbb R$. Suppose that $\hat\mu(u) = \int_\mathbb R e^{iux} d\mu(x) = 0$ for all $u$. Then $\mu(E) = 0$ for all measurable sets $E$.

However, I have not yet met with much success. First of all, is this actually true? And, if so, how can I prove it?

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Most analysis textbooks that discuss Fourier transforms for measures will have a proof of this. I think Folland does, for example. Why not visit your library and then ask here if you have questions about what you see? –  Nate Eldredge Nov 23 '11 at 2:06
    
I just checked Folland, and it does not appear to have a proof of this fact. I also tried searching a couple of different versions of my question on google, and none yielded a site that had a proof. If you could send me a link to a proof somewhere, it would be much appreciated. –  user15464 Nov 23 '11 at 2:24

1 Answer 1

up vote 2 down vote accepted

Here's a sketch of a proof. The details are for you to fill in.

  1. If $f$ is continuous and periodic with some period $M$, then $\int f d\mu = 0$. (Stone-Weierstrass theorem.)

  2. If $f$ is continuous and compactly supported, then $\int f d\mu = 0$. (Approximate $f$ by periodic functions with very long period.)

  3. If you know the Riesz representation theorem, you are done. If not, show that $\mu((a,b))=0$ for all intervals $(a,b)$ by approximating $1_{(a,b)}$ by continuous functions. Then show that $\mu(U) = 0$ for all open sets $U$. Use a monotone class or $\pi$-$\lambda$ argument to show that $\mu(A)=0$ for all Borel sets $A$.

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