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In a paper I currently work with the author wants to show that a complex borel measure on $\mathbb T$ is absolutely continuous. However he only does consider closed null sets of T. Why is this sufficient?

Kind regards, Sebastian

Sebastian Bechtel
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1 Answers1

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Surely the measure $\mu$ you are considering is Radon. Therefore, it has the following "inner regularity" property: if $A\subset \mathbb T$ is any measurable set, then $$\mu(A)=\sup\{\mu(K)\ :\ K\subset A,\ K\ \text{is compact}\}.$$ Since the torus is compact, you can therefore approximate every measurable set with a closed set.

Giuseppe Negro
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  • P.S.: Related. – Giuseppe Negro Sep 14 '15 at 16:28
  • Thanks for your answer! It is not directly given that the measure is radon. And actually, if I understand you correctly, inner regularity is already sufficient. I additionally know that the fourier coefficients of $1$ with respect to that measure are $0$. Does that help? – Sebastian Bechtel Sep 14 '15 at 16:33
  • AFAIK it is easy for a measure to be Radon, especially on a compact space. If its total mass is finite and Borel sets are measurable, then it is Radon. (If you are speaking of Fourier transforms then it must be the case) – Giuseppe Negro Sep 14 '15 at 16:40
  • I think theorem 2.18 in real and complex analysis by Rudin should work. It is formulated for positive measures but I think it should easily extend to complex measures, but so far I didn't check that rigorously. – Sebastian Bechtel Sep 16 '15 at 11:09