Lemma 3.3.11.3FB36EAD186A Let $X$ be a compact metric space and $A\subset X$. Let $\mathcal F$ be a family of closed sets of positive diameter such that every $x\in A$ belongs to members of arbitrarily small diameter. There is a finite or countably infinite disjoint sequence $F_ j\in \mathcal F$ such that, in the finite case, $A\subset \bigcup _ jF_ j$, and, in the infinite case, for every $j$,
\[ A\subset \bigcup _{i\le j}F_ i\ \cup \bigcup _{i>j}N(F_ i,3\operatorname {diam}F_ i). \]
Here $N(F,r)=\{ x:d(x,F)<r\} $. If a finite complete measure $m$ satisfies $0<m(F)$ and $m(N(F,3\operatorname {diam}F))\le C m(F)$ for all eligible members, the sequence covers $A$ except for an $m$-null set.
Proof.
If $A=\varnothing $, choose the empty sequence. Otherwise choose any $F_1$. After $F_1,\ldots ,F_ j$, stop if they cover $A$. If they do not, a point outside their closed union belongs to sufficiently small members $F$ with $N(F,\operatorname {diam}F)$ disjoint from that union. Let $e_ j>0$ be the supremum of their diameters, and choose such an $F_{j+1}$ with diameter $>2e_ j/3$. All chosen sets are disjoint. Fix $x\in A\setminus \bigcup _{i\le j}F_ i$ and choose a member $F$ containing $x$ whose diameter is so small that its diameter-neighborhood avoids the first $j$ sets. It must meet a later chosen $F_ k$ in that neighborhood. Otherwise it remains eligible forever, so all subsequent chosen diameters exceed $2\operatorname {diam}F/3$; choosing a point in each yields an infinite uniformly separated subset of compact $X$, impossible. At the first such $k$, $\operatorname {diam}F_ k>2\operatorname {diam}F/3$. Some point of $F_ k$ is within $\operatorname {diam}F$ of $F$, hence $d(x,F_ k)<2\operatorname {diam}F<3\operatorname {diam}F_ k$. This proves the covering assertion. In the measure case the outer measure of the uncovered set is at most $C\sum _{i>j}m(F_ i)$ for every $j$; the tail tends to zero because the sets are disjoint and $m(X)<\infty $. Completeness makes every subset of the resulting null set measurable.
$\square$
Theorem 3.3.12.5F363A880A82 Let $S=\partial B(0,R)\subset \mathbb R^ d$, $d\ge 2$, with completed surface measure $\sigma $. A finite signed measure $\mu =f\sigma +\mu _ s$ has, for $\sigma $-almost every $\zeta $, the derivative
\[ \lim _{\substack {\zeta \in K\\ \operatorname {diam}K\to 0}} \frac{\mu (K)}{\sigma (K)}=f(\zeta ). \]
The sets $K$ are closed spherical caps of positive diameter less than a hemisphere. Moreover almost every $\zeta $ is a Lebesgue point:
\[ \frac1{\sigma (K)}\int _ K|f(y)-f(\zeta )|\, d\sigma (y)\longrightarrow 0 \quad (\zeta \in K,\ \operatorname {diam}K\to 0). \]
Proof.
Small spherical caps satisfy uniform estimates $c r^{d-1}\le \sigma (K)\le C r^{d-1}$, $r=\operatorname {diam}K$, and $\sigma (N_ S(K,3r))\le C'\sigma (K)$. To verify them, cover $S$ by finitely many graph charts over tangent hyperplanes on which the graph map and its inverse have bounded derivatives and the surface Jacobian is bounded above and below by positive constants. A small cap centered at $z$ contains and is contained in chart balls of radii constant multiples of $r$; its $3r$-neighborhood is contained in a chart ball of radius a constant multiple of $r$. Euclidean volume scaling gives the estimates; compactness supplies uniform constants. These observations also show that any small cap containing $\zeta $ lies in a chart cube centered at $\zeta $ of side $Cr$, while its area is at least $cr^{d-1}$.
The Euclidean Lebesgue differentiation theorem for $L^1_{\rm loc}$ functions on cubes therefore gives the stated absolute-oscillation limit for $f$ in each chart: the numerator is bounded by a constant times the integral of the coordinate oscillation on that cube, since the Jacobian is bounded; the denominator is bounded below by its volume times a positive constant. Null sets in coordinates and on $S$ agree. A finite chart cover proves the assertion almost everywhere. The precise external interface here is the checked cube Lebesgue differentiation theorem lebesgue differentiation of the Foundations source, not a boundary Fatou theorem.
For completeness the singular-measure step follows from the preceding covering lemma rather than an unstated differentiation theorem. If $\nu $ is a finite positive measure and
\[ E_\lambda =\{ \zeta :\limsup _{\zeta \in K,\, \operatorname {diam}K\to 0} \nu (K)/\sigma (K)>\lambda \} ,\qquad \lambda >0, \]
the good caps $\nu (K)>\lambda \sigma (K)$ form a fine cover of $E_\lambda $. The lemma, applied to $m=\sigma $, gives disjoint good caps covering this set up to a surface null set. Thus its outer surface measure is at most $\sum \sigma (K)\le \nu (S)/\lambda $. This argument does not presume measurability of $E_\lambda $. If $\nu $ is singular, choose a compact subset $L$ of a surface-null carrier with $\nu (S\setminus L)<\varepsilon $, by regularity. Outside $L$, every sufficiently small cap sees only $\nu |_{S\setminus L}$. The preceding bound makes the set where the upper density exceeds $\lambda $ have outer area at most $\varepsilon /\lambda $, because $\sigma (L)=0$. Let $\varepsilon \downarrow 0$ and then take rational $\lambda >0$. Hence the upper density is zero almost everywhere. Apply this to $|\mu _ s|$ and combine with the $L^1$ result. The derivative equals measurable $f$ off a null set, so every version defined on the remaining null set is measurable for the completed surface measure.
$\square$
Pome
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