Length of a union of intervals 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 $1$?
 A: 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 $|U|\ge|X|=1$.
A: Yes, $\ m(U)\ge 1$.  The set $\ X\setminus U\ $ is countable, and $\ X\subseteq U\cup(X\setminus U)\ $.  Therefore 
\begin{eqnarray}
1=\ m(X)&\le& m(U \cup (X\setminus U))\\
 &=& m(U) + m(X\setminus U)\\
&=& m(U) + 0\ .
\end{eqnarray}
Proof of countability claim: Let $\ X_n= \left\{ x\in X\setminus U\left\vert\ \epsilon_x > \frac{1}{n}\right.\right\}\ $. If $\ x\in X\setminus U\ $, then since $\ \epsilon_x > 0\ $, there must be some positive integer $\ m\ $ with $\ \frac{1}{m} < \epsilon_x\ $, and so $\ x\in X_m\subseteq\bigcup_\limits{n=1}^\infty X_n\ $.  Therefore $\ X\setminus U=\bigcup_\limits{n=1}^\infty X_n\ $.
But $\ \left\vert X_n\right\vert\ \le n+1\ $, because if $\ x\in X_n\ $ and $\ y\in X_n\ $, with $\ y> x\ $, then $\ y\ge x+\epsilon_x> x+\frac{1}{n}\ $, since otherwise we would have $\ y\in (x, x+\epsilon_x)\subseteq U\ $, whereas $\ y\not\in U\ $ by definition of $\ X_n\ $.  Thus, distinct elements of $\ X_n\ $ are separated by a distance of at least $\ \frac{1}{n}\ $, and all lie in the interval $\ [0,1]\ $, so there can be at most $\ n+1\ $ of them.
Thus, since $\ X\setminus U\ $ is a countable union of finite sets, it is countable.
Acknowledgement: Thanks to zhw for alerting me to a blunder in an earlier attempt to prove this result.
