Vitali covering lemma for Euclidean balls with fivefold dilates

If the finite family is empty, use the empty subfamily. Otherwise retain a ball of maximal radius, breaking ties by least index. Repeat among the remaining balls disjoint from every retained ball, and stop when none remain. There are at most $m$ selections. The retained balls are pairwise disjoint by construction, and the tie rule makes every selection determined.

For the countable family, fix a positive upper bound $R$ for its radii. Let $I_m=\{j\ge 1:2^{-m-1}R<r_j\le 2^{-m}R\}$ for $m\ge 0$; these partition the positive radii. Given the retained indices in earlier classes, inspect $j=1,2,\dots$ and retain $j$ exactly when $j\in I_m$ and its ball is disjoint from every earlier retained ball. Inner recursion on $j$ and outer recursion on $m$ define this rule by [L3], taking the union of each inner sequence before the next class. Every decision is determined, so there is no countable choice of balls. The retained subfamily is pairwise disjoint.

In the finite family, a chosen ball lies in its own fivefold dilate. For an unchosen $B_j$, take the first chosen ball $B_{i_s}$ meeting it. At that stage $B_j$ was eligible, so $r_j\le r_{i_s}$. For $z\in B_j$ and $w\in B_j\cap B_{i_s}$, the centre distance satisfies $\mathrm{\parallel }{c}_j-c_{i_s}{\mathrm{\parallel }}_2\le \mathrm{\parallel }{c}_j-w{\mathrm{\parallel }}_2+\mathrm{\parallel }w-c_{i_s}{\mathrm{\parallel }}_2<r_j+r_{i_s}$. Hence $\mathrm{\parallel }z-c_{i_s}{\mathrm{\parallel }}_2\le \mathrm{\parallel }z-c_j{\mathrm{\parallel }}_2+\mathrm{\parallel }{c}_j-c_{i_s}{\mathrm{\parallel }}_2<2r_j+r_{i_s}\le 3r_{i_s}<5r_{i_s}$. Thus every original ball lies in a chosen fivefold dilate.

Let $B_j$ be any original ball that was not chosen in the countable construction, and let $j\in I_m$. When the algorithm inspected $j$, some previously chosen ball $B_{i_s}$ already met $B_j$; otherwise $B_j$ would have been retained. If $i_s\in I_{\mathrm{\ell }}$ with $\mathrm{\ell }<m$, then $r_j\le 2^{-m}R\le 2^{-\mathrm{\ell }-1}R<r_{i_s}\mathrm{.}$ If instead $\mathrm{\ell }=m$, then both balls lie in the same dyadic class, so $r_j\le 2^{-m}R<2r_{i_s}\mathrm{.}$ In either case, $r_j<2r_{i_s}\mathrm{.}$ Choose $z\in B_j$ and $w\in B_j\cap B_{i_s}$. Then $\mathrm{\parallel }z-c_{i_s}{\mathrm{\parallel }}_2<r_j+r_j+r_{i_s}<2(2r_{i_s})+r_{i_s}=5r_{i_s},$ so again $B_j\subseteq 5B_{i_s}$. Let $I\subseteq {\mathbb{N}}_{\ge 1}$ be the set of retained indices. This set is finite or countably infinite, and chosen balls are also contained in their own fivefold dilates; hence ${\bigcup }_{j\ge 1}{B}_j\subseteq {\bigcup }_{i\in I}5B_i\mathrm{.}$

For the measure consequence, assume countable choice as required by [L2]. Finite subadditivity and the established inclusion give $\lambda ({\bigcup }_{j=1}^m{B}_j)\le \lambda ({\bigcup }_{k=1}^{\mathrm{\ell }}5B_{i_k})\le {\sum }_{k=1}^{\mathrm{\ell }}\lambda (5B_{i_k})=5^n{\sum }_{k=1}^{\mathrm{\ell }}\lambda (B_{i_k})$. This is the only use of countable choice in this proof.

Steps 3.1 and 2.2 prove the finite and countable forms.