Measure with periodic fourier coefficients Rudin asked:
Suppose $\mu$ is a complex Borel measure on $[0,2\pi)$. Define $$\overline{\mu}(n)=\int e^{-int}d\mu$$ Find all $\mu$ such that $\overline{\mu}$ is periodic such that $\mu(n)=\mu(n+k)$ for all $n$. 
I am quite at loss with this problem. Since every continous function can be approximated by a trignometric function, the above condition would effectively imply $$\sum_{j=1}^{n} a_{j}\int e^{-jt}d\mu=0$$ for any linear combination of $a_{i}$ available, for otherwise the periodicity would force we obtain an intergral of a $L_{1}$ function with infinity(consider $f(n)=\sum^{\infty}_{j=1}e^{-jk+n}$), conflicting with the fact that $\mu$ is assumed to be a complex measure. Now let $a_{s}=1,0\le s<k$ and the other coefficients to be $0$, then this would imply $\overline{\mu}(n)=0$ for all $n$. But this condition seems really strong. Did I made any mistake?
On the other hand, if the above condition holds, then $\mu$ seems to be almost everywhere $0$ since we can approximate the characteristic function of any Borel measurable set. This is really surprising for me. I do not know if I made a mistake and Rudin imaged me to attack the problem via other methods. 

If $\widehat{\mu}$ is k-periodic, it can be written as $\widehat{\mu}(n) = \sum_{j=0}^{k-1} c_je^{-2\pi ijn/k}$ for some $c_j\in\mathbb{C}$ . Second, $\widehat{\delta_{j/k}}(n) = e^{-2\pi ijn/k}$, so $$\mu = \sum_{j=0}^{k-1} c_j\ \delta_{j/k}$$ On the other hand, every such $\mu$ has periodic Foruier coefficients.
