Do two closed subsets of $[0, 1]$ with measure $\frac{1}{2}$ intersect?

Let $A$ and $B$ be two closed subsets of $[0,1]$, each with a length of $1/2$. Is it always true that $A\cap B\neq \emptyset$?

My intuition is yes, because:

• Either they intersect in their interior;
• Or, they are interior-disjoint (i.e. $\operatorname{int}(A) = [0,1]\setminus B$), but in this case they will intersect at their boundary.

What is a formal proof to this claim?

Also, I will be thankful for references that discuss possible generalizations of this claim to more than two subsets (possibly in $\mathbb{R}^n$).

Let $A$, $B$ be two closed subsets of $[0, 1]$, both with measure $\frac{1}{2}$, and suppose $A\cap B = \emptyset$. Then

$$m([0, 1]\setminus(A\cup B)) = m([0, 1]) - (m(A) + m(B)) = 1 - \left(\tfrac{1}{2} + \tfrac{1}{2}\right) = 0.$$

But $[0, 1]\setminus(A\cup B)$ is open and the only open set with measure zero is the empty set, so $[0, 1]\setminus(A\cup B) = \emptyset$; i.e. $A\cup B = [0, 1]$.

As $A$ and $B$ are two non-empty disjoint closed sets with union $[0, 1]$, $A^c$ and $B^c$ are two non-empty disjoint open sets with union $[0, 1]$. But this is a contradiction as $[0, 1]$ is connected. Therefore, $A\cap B \neq \emptyset$.

The above argument can easily be adapted to prove the following generalisation: for any two closed subsets $A$, $B$ of $[0, 1]$ with $m(A) + m(B) = 1$, $A\cap B \neq \emptyset$.

• I like your proof! +1 – Nicolas May 3 '15 at 9:30

Also, I will be thankful for references that discuss possible generalizations of this claim to more than two subsets and to $\mathbb{R}^n$.

It seems the following.

Proposition. Let $K\subset\Bbb R^n$ be a convex body (that is, the interior of the set $K$ is non-empty), $K_1,\dots, K_n$ be a finite family of closed subsets of the set $K$ such that $\sum\mu(K_i)\ge\mu(K)$. There there exist different $K_i$ and $K_j$ with non-empty intersection.

Proof. Assume the converse. Since the set $K$ is connected, the set $\Delta=K\setminus\bigcup K_i$ is non-empty. Since $\Delta$ is an open subset of the convex body, $\mu(\Delta)>0$, a contradiction.

A generalization for the countable number of closed subsets fails already for the unit segment.

• Don't you have to assume that the sets $K_1,...,K_n$ are closed? – Erel Segal-Halevi May 3 '15 at 17:30
• @ErelSegal-Halevi Yes. I corrected this misprint. – Alex Ravsky May 3 '15 at 17:47