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Prove that there is no Lebesgue measurable set $A$ in [0,1] with the property that $m(A\cap I) = \frac{1}{4}m(I)$ for every interval $I$

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This seems to be essentially the same as this question. – user51696 Dec 4 '12 at 8:05
Dear @dan, if you registered your account then you could (eventually) vote instead of just posting comments. That would be much tidier. Let me know what you think. – Rudy the Reindeer Dec 4 '12 at 11:02
up vote 3 down vote accepted

Let $U$ be an open set of $\mathbb{R}$ such that $A \subset U$ and $m(U\setminus A) \leq \epsilon$. Being an open set of $\mathbb{R}$, $U$ can be written as a countable (or finite) disjoint union $$ U = \bigcup_{n} (a_n,b_n). $$

Therefore, we obtain $$ m(A) = m(A \cap U) \leq \sum_n m(A\cap (a_n,b_n)) = \frac{1}{4} \sum_n (b_n-a_n) = \frac{1}{4}m(U) \leq \frac{1}{4}(m(A) + \epsilon), $$

hence $\forall \epsilon > 0$, $m(A) \leq \epsilon/3$. Finally, $m(A)=0$ and $m(A \cap [0,1]) = 0 \neq 1/4$.

In the exact same way, we have

Lemma. If $A$ is a Borel subset of $\mathbb{R}$ such that there exists $\lambda < 1$ with $$\forall a,b \in\mathbb{R} ,\quad a < b \implies m(A \cap (a,b)) \leq \lambda(b-a)$$ then $A$ is negligible.

Alternative proof using Dynkin's lemma.

Suppose that $A$ exists. Then the set of Borel subsets $B$ of $[0,1]$ such that $m(A\cap B) = \frac{1}{4}m(B)$ is a $\lambda$-system which contains the $\pi$-system of all intervals, hence the Borel $\sigma$-algebra of $[0,1]$. In particular, $m(A)=m(A \cap A)=\frac{1}{4}m(A)$ and $m(A)=m(A\cap[0,1])=\frac{1}{4}$, which is absurd.

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You have to add that $U$ is a disjoint union of open intervals. (the argument is correct, so I upvote) – Davide Giraudo Dec 4 '12 at 10:25
Thanks. Actually, this is not critical here since we are only interested in inequalities. – Siméon Dec 4 '12 at 10:50
Thanks. In fact there is a hint along this problem that suggests to consider the characteristic function of $A$. – Hongshan Li Dec 5 '12 at 0:19
Maybe the author of the problem was thinking of the functional version of Dynkin's lemma. Here it would give you that for all non-negative mesurable $f$,$$\int 1_Af\,dm = \frac{1}{4}\int f\,dm$$ which leads to a contradiction (see the alternative proof above). – Siméon Dec 5 '12 at 10:01

Hint: Lebesgue's density theorem ('s_density_theorem ) states that if $A\subset\mathbb{R}$ is Lebesgue-measurable, then for almost every $x\in\mathbb{R}$ we have \begin{equation*} \lim_{r\to 0}\frac{m(A\cap B(x,r))}{m(B(x,r))}=1, \end{equation*} where $B(x,r)$ is the open $r$-radius ball centered at $x$. In $\mathbb{R}$ these are open intervals.

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The following property of Lebesgue measurable sets is useful here: Given a measurable set $A$, there exists sets $F,U$ such that $F$ is closed, $U$ is open, and $F\subset A\subset U$, and most importantly $m(U\setminus F)<\epsilon$ for any $\epsilon$.

From this proposition, you can prove without too much extra work that every Lebesgue measurable set of finite measure greater than zero almost contains a full interval.

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What does containing almost a full interval mean? If we take for example $A=[-2,-1]\cup[1,2]$, then what would it mean in this case? – T. Eskin Dec 4 '12 at 7:45
Precisely, it means given $\epsilon>0$, there exists an interval $I$ such that $m(I\cap A)\geq (1-\epsilon)m(I)$. – rondo9 Dec 4 '12 at 14:22

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