TITLE: Do two closed subsets of $[0, 1]$ with measure $\frac{1}{2}$ intersect?
QUESTION [8 upvotes]: 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 $\mathbb{R}^n$).

REPLY [3 votes]: 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.