# If the Fourier transform of a signed measure is identically zero, is the same true of the measure?

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

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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How is the 1.) the case? Don't $\sin(x)$ and $\cos(x)$ have these properties, but their integrals are not zero? – jablesauce Apr 3 '15 at 22:54
@jablesauce Note we are working under the hypothesis that $\mu$ is a measure such that $\int e^{iux} \mu(dx) = 0$ for all $u$. Then $\int \sin(x) \mu(dx) = \frac{1}{2i}\left(\int e^{ix}\mu(dx) + \int e^{-ix} \mu(dx)\right) = 0$ applying our assumption with $u=1$ and $u=-1$. So indeed, for such a measure $\mu$ we do have that the integral of $\sin(x)$ and $\cos(x)$ are zero. (Since it turns out that $\mu$ is the zero measure, this is not surprising.) – Nate Eldredge Apr 3 '15 at 23:14
Understood, thanks! What did you mean exactly when you wrote $u=1$ and $u=−1$ ? – jablesauce Apr 3 '15 at 23:21
@jablesauce: When you apply $\int e^{iux} \mu(dx) = 0$ with $u=1$ it says $\int e^{ix} \mu(dx) = 0$, and when you apply it with $u=-1$ it says $\int e^{-ix} \mu(dx) = 0$. – Nate Eldredge Apr 3 '15 at 23:25
Sorry my mistake and didn't notice the $u$ in the integral in your previous comment. Thanks again! – jablesauce Apr 3 '15 at 23:45