TITLE: If the Fourier transform of a signed measure is identically zero, is the same true of the measure?
QUESTION [6 upvotes]: 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?

REPLY [7 votes]: Here's a sketch of a proof.  The details are for you to fill in.

If $f$ is continuous and periodic with some period $M$, then $\int f d\mu = 0$.  (Stone-Weierstrass theorem.)
If $f$ is continuous and compactly supported, then $\int f d\mu = 0$.  (Approximate $f$ by periodic functions with very long period.)
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 $A$.