I am reading the book Vector Measures of Diestel and Uhl, especifically example 6 of Sierpinski in Chapter 2, about the construction of a function that is weak$^*$-measurable but not weakly measurable. As part of the construction, the authors find a function $\phi$ on $[0,1]$ that only takes values 0 and 1, except for the countable set of dyadic numbers. They also show that $\phi$ has a dense set of periods, especifically, that $\phi(t+d)=\phi(t)$ for every $t, t+d \in[0,1]$ with $d$ dyadic. If $\phi$ were Lebesgue measurable they conclude that $\phi$ is a constant $k$ almost everywhere with respect to Lebesgue measure, but I cannot see why. For all I see, a constant $k$ repeats at least countable times, but why almost everywhere? Any help explaining this will be very appreciated.
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$\begingroup$ The function is only defined on $[0,1]$, so $t+d$ outside that is excluded. $\endgroup$– S.O.C.Commented Sep 22, 2020 at 23:14
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$\begingroup$ I have also tried using that the measure of $\phi^{-1}(\{0\})$ is positive or the measure of $\phi^{-1}(\{1\})$ is positive, but I still find no clarity as to why the measure of one of those is 1. $\endgroup$– S.O.C.Commented Sep 22, 2020 at 23:18
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$\begingroup$ I am also stuck using outer regularity and inner regularity of Lebesgue measure. $\endgroup$– S.O.C.Commented Sep 23, 2020 at 0:48
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$\begingroup$ This looks suspicious under these assumptions. What about starting with the standard example of a non-measurable set, which has the property that its rational translates partition $\mathbb R$, and then taking $\phi^{-1}(1)$ as its dyadic translates only. This set feels non-measurable also. Maybe $\phi$ has additional properties that you haven't mentioned? $\endgroup$– Christian RemlingCommented Sep 23, 2020 at 1:12
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2$\begingroup$ In this version, this is a quick consequence of Lebesgue's density theorem: let $A=\phi^{-1}(1)$. Almost every point of a measurable set is a point of density. If we had $0<|A|<1$, then both $A$ and $A^c$ have points of density, but this can't be true here because we can take a part of $A$ that almost fills a small interval and use dyadic translates to maneuver this to a point of density of $A^c$. $\endgroup$– Christian RemlingCommented Sep 23, 2020 at 1:58
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