The Vitali covering theorem for fine covers on the real line

Choose an open set $U\supseteq E$ with finite length and $\lambda (U)<{\lambda }^{\ast }(E)+1$. Because $\mathcal{V}$ is fine, every $x\in E$ lies in arbitrarily short intervals of $\mathcal{V}$, so after shrinking we may restrict to the subfamily ${\mathcal{V}}_U:=\{I\in \mathcal{V}:I\subseteq U\},$ which is still a fine cover of $E$. Using dependent choice, choose a sequence $(I_n{)}_{n\ge 1}$ in ${\mathcal{V}}_U\cup \{\mathrm{\varnothing }\}$ as follows: as long as there exists an interval of ${\mathcal{V}}_U$ disjoint from the previously chosen nonempty intervals, choose $I_n$ disjoint from those earlier nonempty intervals and with length at least half the supremum of the lengths of all such candidates; once no disjoint candidate remains, set $I_n=\mathrm{\varnothing }$ and all later terms equal to $\mathrm{\varnothing }$.

Let $M:=\{n\ge 1:I_n\mathrm{\ne }\mathrm{\varnothing }\}$. The nonempty intervals $(I_n{)}_{n\in M}$ are pairwise disjoint and lie in the finite-open set $U$, so ${\sum }_{n\in M}\mathrm{\mid }{I}_n\mathrm{\mid }\le \lambda (U)<\mathrm{\infty }\mathrm{.}$ If $M$ is finite, write $M=\{1,\dots ,N\}$ and let $R:=E\setminus {\bigcup }_{n=1}^N{I}_n$. If $x\in R$, then the finite union ${\bigcup }_{n=1}^N{I}_n$ is closed and does not contain $x$, so $d:=\mathrm{dist}(x,{\bigcup }_{n=1}^N{I}_n)>0$. Because ${\mathcal{V}}_U$ is a fine cover of $E$, choose $J\in {\mathcal{V}}_U$ with $x\in J$ and $\mathrm{\mid }J\mathrm{\mid }<d$. Then $J$ is disjoint from $I_1,\dots ,I_N$, contradicting the terminal clause in step 1.1. Therefore $R=\mathrm{\varnothing }$. In this finite-termination case the theorem already holds with the finite family $I_1,\dots ,I_N$, and the finite $\epsilon$-version is immediate as well.

It remains to treat the case where $M$ is infinite. Then no $I_n$ is empty, because empties persist forever after the first one, so ${\sum }_{n\ge 1}\mathrm{\mid }{I}_n\mathrm{\mid }<\mathrm{\infty }$ and therefore $\mathrm{\mid }{I}_n\mathrm{\mid }\to 0$. Let $R:=E\setminus {\bigcup }_{n\ge 1}{I}_n$. Fix $x\in R$ and $m\ge 1$. Since $x$ does not lie in the finite union $I_1\cup \cdots \cup I_{m-1}$, the distance $d_m:=\mathrm{dist}\text{\hspace{0.17em}⁣}(x,{\bigcup }_{n=1}^{m-1}{I}_n)$ is positive. Also $s_m:={\sup }_{n\ge m}\mathrm{\mid }{I}_n\mathrm{\mid }>0,$ because there are infinitely many nonempty intervals in the tail. Choose $J\in {\mathcal{V}}_U$ with $x\in J,\mathrm{\mid }J\mathrm{\mid }<\min (d_m,s_m)\mathrm{.}$ Then $J$ is disjoint from $I_1,\dots ,I_{m-1}$. If $J$ were disjoint from every $I_n$ with $n\ge m$, then $J$ would remain a candidate at every later stage. The half-maximal choice would then give $\mathrm{\mid }J\mathrm{\mid }\le 2\mathrm{\mid }{I}_n\mathrm{\mid }$ for every $n\ge m$, contradicting $\mathrm{\mid }{I}_n\mathrm{\mid }\to 0$. So $J$ meets some $I_n$ with $n\ge m$; let $n$ be the least such index. Then $J$ is disjoint from $I_1,\dots ,I_{n-1}$, so it is an admissible candidate at stage $n$, and the half-maximal choice gives $\mathrm{\mid }J\mathrm{\mid }\le 2\mathrm{\mid }{I}_n\mathrm{\mid }$. Intersecting intervals with comparable lengths satisfy $J\subseteq 5I_n$. Therefore $x\in {\bigcup }_{n\ge m}5I_n$. Since $m$ was arbitrary, $R\subseteq {\bigcap }_{m\ge 1}{\bigcup }_{n\ge m}5I_n\mathrm{.}$

By countable subadditivity of outer measure and the interval formula for Lebesgue outer measure, ${\lambda }^{\ast }(R)\le {\inf }_{m\ge 1}{\lambda }^{\ast }\text{\hspace{0.17em}⁣}\left({\bigcup }_{n\ge m}5I_n\right)\le {\inf }_{m\ge 1}5{\sum }_{n\ge m}\mathrm{\mid }{I}_n\mathrm{\mid }=0.$ This proves the countable disjoint conclusion in the infinite case. For the finite version, choose $N$ so large that $5{\sum }_{n>N}\mathrm{\mid }{I}_n\mathrm{\mid }<\epsilon$ and repeat the same estimate with $R_N:=E\setminus {\bigcup }_{n=1}^N{I}_n$.

Steps 1.1 through 4.1 prove both forms of the theorem.