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If $f$ is analytic in $\{z\in \mathbb{C}, \Im z>0\}$, and continuous in $\{z\in \mathbb{C}, \Im z\ge 0\}$. I'm curious about the structure of the set $$ E=\{z\in \mathbb{R},~~ f(z)=0\} $$ When restrict $f$ on the real line, it's not analytic, so $E$ may not be discrete.

My question: Is there an example such that the set $E$ contains uncountable points (such as a cantor set)?

Is it true that $E\subset \mathbb{R}$ is of measure $0$?

Tomas
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  • Remark. Can we get an example with a dense set of zeros on the boundary? The set of zeros of a continuous function must be a $G_\delta$-set, so if it is dense then it is uncountable by the Baire category theorem. – GEdgar Dec 01 '15 at 15:59
  • @GEdgar No, the zero set must be closed and nowhere dense. – Daniel Fischer Dec 01 '15 at 21:11
  • Yes, exactly the closed measure zero sets $E$ are possible as zero sets. This is classical, but requires some machinery. See for example here: http://math.stackexchange.com/questions/6208/isolated-zeros-on-closure-of-a-domain –  Dec 02 '15 at 03:52

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