Vitali covering lemma for Euclidean balls with fivefold dilates

For the finite family, choose $B_{i_1}$ with maximal radius among [given, choose] $B_1,\dots ,B_m$. Having chosen disjoint balls $B_{i_1},\dots ,B_{i_{s-1}}$, choose $B_{i_s}$ with maximal radius among the remaining balls disjoint from all earlier choices, and stop when none remain. The chosen subfamily is pairwise disjoint by construction.

Now let $(B_j{)}_{j\ge 1}$ be countable with radii bounded above, and put $R:={\sup }_{j\ge 1}{r}_j\mathrm{.}$ For each integer $m\ge 0$, let $I_m:=\{j\ge 1:2^{-m-1}R<r_j\le 2^{-m}R\}\mathrm{.}$ Process the classes $I_0,I_1,\dots$ in this order, and within each class inspect the indices in increasing order. Retain $B_j$ exactly when it is disjoint from every ball already retained. The retained subfamily is pairwise disjoint by construction.

Let $B_j$ be one of the original balls. If it was chosen, then [step 1.1, L1, choose, algebra] $B_j\subseteq 5B_j$. If it was not chosen, let $B_{i_s}$ be the first chosen ball that meets it. Since the choice at stage $s$ had maximal radius among the remaining disjoint balls, the radius of $B_j$ is at most that of $B_{i_s}$. Pick $z\in B_j$ and choose $w\in B_j\cap B_{i_s}$. Then $\mathrm{\parallel }z-c_{i_s}{\mathrm{\parallel }}_2\le \mathrm{\parallel }z-w{\mathrm{\parallel }}_2+\mathrm{\parallel }w-c_j{\mathrm{\parallel }}_2+\mathrm{\parallel }{c}_j-c_{i_s}{\mathrm{\parallel }}_2<r_j+r_j+r_{i_s}\le 5r_{i_s},$ so $z\in 5B_{i_s}$. Therefore every original ball lies in the union of the fivefold dilates of the chosen balls.

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 ${\bigcup }_{j\ge 1}{B}_j\subseteq {\bigcup }_{i\in I}5B_i\mathrm{.}$

Since the chosen balls are pairwise disjoint, [step 2.1, L2, algebra] $\lambda \text{\hspace{0.17em}⁣}\left({\bigcup }_{j=1}^m{B}_j\right)\le \lambda \text{\hspace{0.17em}⁣}\left({\bigcup }_{k=1}^{\mathrm{\ell }}5B_{i_k}\right)\le {\sum }_{k=1}^{\mathrm{\ell }}\lambda (5B_{i_k})=5^n{\sum }_{k=1}^{\mathrm{\ell }}\lambda (B_{i_k})\mathrm{.}$ This proves part 1.

Steps 3.1 and 2.2 prove the finite and countable forms.