# A Lebesgue measure question [duplicate]

Is there a measurable set $E \subset [0,1]$ such that for any $0 < a < b<1$, the Lebesgue measure $$m(E \cap [a,b])= \frac{b-a}{2}?$$

I am stumped and have no idea.

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## marked as duplicate by GEdgar, JavaMan, t.b., Zhen Lin, Dylan MorelandDec 8 '11 at 6:18

I am learning real analysis from an old grad level book I got, and this is just one of the problems in it. So maybe a homework problem? – Alex J. Dec 8 '11 at 0:14
To make the linked question more directly applicable, note that an open subset $U$ of $[0,1]$ is a countable union of disjoint open intervals, and it follows that $m(E\cap U)=\frac{1}{2}\cdot m(U)$. (And then you can take $U$ such that $E\subset U$ and $m(U)<1$ to get a contradiction.) – Jonas Meyer Dec 8 '11 at 6:27

No, there is no such a set.

Since $E$ is measurable (with positive measure) for all $\epsilon > 0$ there exist a sequence of intervals $(I_n)$ in $[0,1]$ such that $E \subset \bigcup I_n$ and $$\sum m(I_n) < m(E) + \epsilon.$$ Hence we obtain that $$\frac{3}{4} \sum m(I_n) < m(E).$$ Thus $$\frac{3}{4} \sum m(I_n) < m(E) \leq m(E \cap \bigcup I_n) = m(\bigcup E \cap I_n) \leq \sum m(E \cap I_n),$$ which implies that there exists positive integer $k$ such that $$\frac{3}{4} m(I_k) \leq m(E \cap I_k).$$ The last inequality contradicts your assumption that for any interval $I$ we have $m(E \cap I) = m(I)/2$.

Whats more, in the exactly same way you can prove general fact:

If $E \subset \mathbb{R}$ is a measurable set with positive measure and $0 \leq c < 1$ than there exists an interval $I$ such that $$m(E \cap I) \geq c \cdot m(I).$$

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No there can be no such $E$.

Let $\chi_E$ be the characteristic function of $E$. At every Lebesgue point $x$ of $\chi_E$, we have

$$\chi_E(x) = \lim_{r\to 0}\ \frac{1}{m(B_r)} \int_{B_r(x)} \chi_E(y) \ dy$$

Since almost every point of an integrable function is a Lebesgue point, the set of $x$ for which this limit is not equal to $0$ or $1$ must have measure $0$.

In particular, it cannot be true that

$$m(E\cap[a,b]) = \frac{b-a}{2}$$

for all $a,b$, since that would imply that also

$$\frac{1}{2} = \frac{m(E\cap[x-r,x+r])}{m([x-r,x+r])} = \frac{1}{m(B_r)} \int_{B_r(x)} \chi_E(y) \ dy$$

for all $x$ and all small enough $r>0$. (Which would be in contradiction to the fact about Lebesgue points above.)

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