= Solution
Start with a finite subcover of the <compact set> $K$. Among its remaining <open balls>, retain one of greatest radius and discard every ball intersecting it. Repeat until no balls remain. The retained family $\mathcal F$ is finite and pairwise disjoint. If a discarded ball $B(c,r)$ met the retained ball $B(c_0,R)$, then $r\le R$ and $|c-c_0|<r+R$. Therefore, for any $y\in B(c,r)$,
$$
|y-c_0|\le |y-c|+|c-c_0|<2r+R\le3R.
$$
Every original ball lies in the concentric triple of one retained ball. This proves the <Wiener covering lemma> for the finite subcover. By <Lebesgue measure> scaling and subadditivity,
$$
\lambda_d(K)\le\sum_{B\in\mathcal F}\lambda_d(3B)
=3^d\sum_{B\in\mathcal F}\lambda_d(B).
$$
Disjointness gives
$$
\boxed{\lambda_d\left(\bigcup_{B\in\mathcal F}B\right)
\ge3^{-d}\lambda_d(K).}
$$
The empty <compact set> is covered by the empty selection.
The <spherical derivative of a measure> at $x$ is the limit
$$
D_s\mu(x)=\lim_{r\downarrow0}
\frac{\mu(B(x,r))}{\lambda_d(B(x,r))}
=\lim_{r\downarrow0}\frac{\mu(B(x,r))}{v_dr^d},
$$
when it exists, where $v_d$ is the unit-ball volume. We prove that this limit is zero outside a Lebesgue-<null set> for the singular <measure>.
First derive the needed maximal estimate from the selection lemma. For a finite positive <Borel measure> $\eta$, put
$$
M_u\eta(x)=\sup_{\substack{B\text{ an open ball}\\x\in B}}
\frac{\eta(B)}{\lambda_d(B)}.
$$
This is the <uncentered maximal function of a finite measure>. For $\alpha>0$, its strict superlevel set is the union of all balls with $\eta(B)>\alpha\lambda_d(B)$, so it is open. Cover any <compact subset> $H$ of this set by such balls and apply the finite <Wiener covering lemma>. The selected disjoint balls give
$$
\lambda_d(H)\le3^d\sum_{B\in\mathcal F}\lambda_d(B)
\le\frac{3^d}{\alpha}\sum_{B\in\mathcal F}\eta(B)
\le\frac{3^d}{\alpha}\eta(\mathbb R^d).
$$
By <inner regularity of Lebesgue measure>, this proves the <uncentered maximal weak-type inequality>
$$
\boxed{\lambda_d\{M_u\eta>\alpha\}
\le\frac{3^d}{\alpha}\eta(\mathbb R^d).}
$$
Since $\mu$ and $\lambda_d$ are <mutually singular measures>, there is a Borel set $N$ with $\lambda_d(N)=0$ and $\mu(\mathbb R^d\setminus N)=0$. By <regularity of finite Borel measures on Euclidean space>, for each $\varepsilon>0$ choose compact $L\subset N$ with $\mu(\mathbb R^d\setminus L)<\varepsilon$. Define the remainder <measure> $\eta_\varepsilon(A)=\mu(A\setminus L)$.
For $x\notin L$, distance from $x$ to this <compact set> is positive. Thus sufficiently small centered balls avoid $L$, and
$$
\mu(B(x,r))=\eta_\varepsilon(B(x,r))
\quad\text{for all sufficiently small }r.
$$
Write $\overline D_s\mu(x)$ for the upper limit of the ratios. If it exceeds $\alpha$, then either $x\in L$ or $M_u\eta_\varepsilon(x)>\alpha$. Hence
$$
\{\overline D_s\mu>\alpha\}
\subset L\cup\{M_u\eta_\varepsilon>\alpha\}.
$$
The right side has <Lebesgue measure> at most $3^d\varepsilon/\alpha$, since $\lambda_d(L)=0$. This even bounds the <outer measure> of the left side, without a separate measurability argument for the upper density. Let $\varepsilon\downarrow0$, and then take the countable union over positive rational $\alpha$. We obtain $\overline D_s\mu=0$ Lebesgue almost everywhere. The ratios are nonnegative, so their lower limit is also zero and
$$
\boxed{D_s\mu(x)=0\qquad\lambda_d\text{-almost everywhere}.}
$$
This proves that the <spherical derivative of a singular measure vanishes>. It is important to approximate the null carrier by <compact subsets>: a null carrier can be dense, so points outside it need not have any neighbourhood avoiding it.
In one dimension, consider the <cumulative distribution function> $F$. For either sign of $h\ne0$, monotonicity gives a nonnegative difference quotient, and the relevant half-open interval lies inside $B(x,2|h|)$. In detail, the numerator is $\mu((x,x+h])$ for $h>0$, while after reversing both signs it is $\mu((x+h,x])$ for $h<0$. Thus
$$
0\le\frac{F(x+h)-F(x)}h
\le\frac{\mu(B(x,2|h|))}{|h|}
=4\,\frac{\mu(B(x,2|h|))}{\lambda_1(B(x,2|h|))}.
$$
At each point where the <spherical derivative of a measure> is zero, this bound tends to zero from both sides. The enlarged open interval avoids any endpoint ambiguity from atoms. Therefore \b[the ordinary two-sided <derivative> exists and]
$$
\boxed{F'(x)=0\qquad\lambda_1\text{-almost everywhere}.}
$$
This is why <singular distribution functions have zero derivative almost everywhere> without having to be constant: almost-everywhere differentiation recovers increments only under extra hypotheses such as <absolute continuity of a function>.
Back to article page
