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Is there a measurable set $A$ such that $m(A \cap B)= \frac12 m(B)$ for every open set $B$?

Edit: (t.b.) See also A Lebesgue measure question for further answers.

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Hint: Lebesgue density theorem.

Alternatively, approximate $A\cap[0,1]$ with a finite union of intervals.


On second thought, those hints are overly complicated. You can use the definition of Lebesgue measure to find an open set $B$ containing $A\cap[0,1]$ with measure close to that of $A\cap[0,1]$.

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  • $\begingroup$ thanks Jonas, but I quickly looked up Lebesgue density theorem but wasnt sure how to apply it. as your other suggestion. i know m(A ∩ [0,1])=1/2, im not sure what you are trying to say by approximate it with finite union of intervals $\endgroup$ – jack Feb 3 '11 at 2:07
  • $\begingroup$ For Lebesgue density, think about what it says for the intersection of $A$ with small intervals. For the second hint, for every set of finite measure $E$ (here $E=A\cap[0,1]$) and every $\varepsilon>0$ there is a finite union of intervals $U$ such that the measure of the symmetric difference of $E$ and $U$ is less than $\varepsilon$. $\endgroup$ – Jonas Meyer Feb 3 '11 at 2:28

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