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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.

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I would begin with $\mu(n+k)-\mu(n) = \int e^{-int} \left(e^{-ikt}-1\right)\,d\mu$. Hence we want the measure $\nu = \left(e^{-ikt}-1\right)\,\mu$ to satisfy $\int e^{-int} \,d\nu=0$ ... hm, are you sure this is really for all integers $n$, or just for nonnegative ones? –  user53153 Jan 3 '13 at 3:22
all integers $n$. Background: Real and Complex Analysis, 1st edition, Chapter 6, problem 8. –  Bombyx mori Jan 3 '13 at 3:25
OK, then we are done: $\nu$ must be zero. Can you see what exactly this means for $\mu$, and if yes, will you post an answer below? –  user53153 Jan 3 '13 at 3:26
I agree $v$ must be 0 almost everywhere. But if $\mu$ is not 0, then we would have trouble since by polar decomposition we have $dv=(e^{-ikt}-1)d\mu=h(e^{-ikt}-1)|d\mu|$. Since $h$ and $e^{-ikt}$ are both not 0, we have shown $\mu$ must be 0 almost everywhere. –  Bombyx mori Jan 3 '13 at 3:32
Yes, I would be happy to post an answer along these lines. But I not unclear if my original solution is correct. Let me think. –  Bombyx mori Jan 3 '13 at 3:34
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