Length of a union of intervals

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 $|U|\ge|X|=1$.