Construct a Borel subset $A$ of $\mathbb{R}$ such that $0 < m(A \cap I) < m(I)$ for every open interval $I$.

Asked Sep 11 '18 at 17:15

Active Sep 11 '18 at 17:15

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Let $m$ be Lebesgue measure. Construct a Borel subset $A$ of $\mathbb{R}$ such that $0 < m(A \cap I) < m(I)$ for every open interval $I$.

This question has been discussed before on this site, for example, here.

However, I came up with a much simpler solution than the other answers on this site, so I am worried there is something wrong with it.

Suppose $I$ is an open interval. For any set $A$, we define the set $cA = \{ca : a \in A \}$. Let $A = \frac{1}{2}I$. The set $A$ is also an interval and is therefore a Borel set. Since $m(cA) = \lvert c \rvert m(A)$, we have that $m(A \cap I) = m(\frac{1}{2} I) = \frac{1}{2} m(I)$. Then we have $$ 0 < m(A \cap I) = \frac{1}{2} m(I) < m(I) $$ as required.

Is there something wrong with this proof?

asked Sep 11 '18 at 17:15

measuresproblem

3

It only works for one interval. – Angina Seng Sep 11 '18 at 17:17
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@LordSharktheUnknown I see! Thank you! I knew it was something simple. :( – measuresproblem Sep 11 '18 at 17:18
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In addition, if $I=[4,6]$, then $A=\frac12I=[2,3]$, hence $A\cap I=\emptyset$, also in violation of your requirement. But the main point, as Lord Shark said, is to find a set that works for any interval. – Ingix Sep 11 '18 at 17:43

Construct a Borel subset $A$ of $\mathbb{R}$ such that $0 < m(A \cap I) < m(I)$ for every open interval $I$.

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