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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