TITLE: Length of a union of intervals
QUESTION [13 upvotes]: Let $X$ be a subset of $[0,1]$ with length (Lebesgue measure) $1$.
For each $x\in X$, there is some $\epsilon_x > 0$. Define $I_x$ as the open interval $(x, x+\epsilon_x)$. I am interested in the union of all these intervals:
$$U := \bigcup_{x\in X} I_x$$
Initially I thought that $U$ contains the entire unit interval, but this is not true. For example, it is possible that for each $x<0.5$, $\epsilon_X := (0.5 - x)/2$. Then, $0.5 \not \in U$.
My quesion is: is the length of $U$ always at least $1$?

REPLY [5 votes]: Let $A=X\setminus U$. We want to show that $|A|=0$, where $|\cdot|$ denotes the Lebesgue measure of any measurable set. Suppose by contradiction that $|A|>0$. 
By Lebesgue's density theorem,
for any measurable set $A\subset\mathbb{R}$, the $1$-dimensional density $\Theta^1(x,A)$ exists at almost every point $x$ and equals $1$ at almost every point of $A$. Since $|A|>0$, there is a point $x\in A$ such that $\Theta^1(x,A)$ exists and equals $1$, that is
$$ \lim_{r\searrow 0} \frac{|A\cap (x-r,x+r)|}{2r}=1. $$
Since $x\in X$, $(x,x+\epsilon_x)\subset U$ for some $\epsilon_x>0$, and thus $(x,x+\epsilon_x)\cap A=\emptyset$. It follows that for any $0<r<\epsilon_x$ we have that
$$ |A\cap (x-r,x+r)|\le |(x-r,x)|=r, $$
and then
$$ \lim_{r\searrow 0} \frac{|A\cap (x-r,x+r)|}{2r}\le\frac12, $$
that gives a contradiction. Hence we have that $|A|=|X\setminus U|=0$, that is to say that almost every point of $X$ belongs to $U$, and thus $|U|\ge|X|=1$.