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A set $E$ is measurable for an outer measure ${\mu }^{\ast }$ when every set $A$ is split additively by $E$:The measurable sets form a sigma-algebra, and the restriction of ${\mu }^{\ast }$ to it is a complete measure.

For measurable sets $E_n$, a measure satisfiesReplace $E_n$ by the disjoint sets $E_n\setminus {\bigcup }_{j<n}{E}_j$, use countable additivity, and then monotonicity. The same property is an axiom for an outer measure on arbitrary sets.

Hausdorff measure measures the geometric size of a set using covers by increasingly small pieces. The exponent $s$ chooses the measuring scale: $s=1$ is length-like, $s=2$ is area-like, and noninteger exponents can measure fractals. Unlike a count of occupied grid cells, the construction allows pieces of different sizes and considers every admissible cover.

For $s>0$ and a subset $E$ of a metric space, use the diameter-cost conventionThe first quantity is scale-restricted Hausdorff content. Shrinking $\delta$ removes options from the infimum, so these contents increase to a limit, possibly infinite. The resulting outer measure restricts to a measure on the Borel sets. For $s=0$, counting each nonempty covering piece with cost one gives counting measure.

A different common convention multiplies these quantities by $c_s={\pi }^{s/2}/(2^s\Gamma (1+s/2))$, using the Gamma function. At integer $s$, that normalization agrees with ordinary Euclidean $s$-dimensional volume. Here $c_1=1$, so both conventions make ${\mathcal{H}}^1$ agree with arc length on smooth curves in Euclidean space. Positive normalization constants never change Hausdorff dimension. A definition must specify this constant when numerical values of Hausdorff measure matter.

For a line segment of length $L$, splitting it into $n$ equal pieces gives the cover costIt tends to zero for $s>1$, stays at $L$ for $s=1$, and grows for $s<1$. The matching lower bound comes from the fact that any interval cover must have total length at least $L$. The exponent one is therefore the transition between a measuring rule too coarse to see the segment and one that assigns it infinite size. For a Cantor set, the transition is instead at $\log 2/\log 3$, as shown in Hausdorff dimension of the Cantor set.

This makes Hausdorff measure useful when ordinary Lebesgue measure sees too little: a curve in the plane has zero planar area but can have positive arc length, and the Cantor set has zero line length but positive geometric size at its own dimension. It also separates local covering estimates from global cardinality: a countable dense set has zero Hausdorff dimension, despite meeting every open region of its closure.

An original source for the construction is Felix Hausdorff's ^[ref]. The modern covering and Hausdorff dimension framework is developed in Bishop and Peres, Fractals in Probability and Analysis, §1.2.

The mass distribution principle converts a diffuse measure into a lower bound on Hausdorff dimension. Let $s>0$ and let a Borel probability measure $\mu$ give a Borel set $E$ full mass. If, for some $C,r_0>0$, every ball centered in $E$ satisfiesthen ${\mathcal{H}}^s(E)\ge 1/C>0$ in the diameter-cost convention, and consequently ${\dim }_{H}E\ge s$.

The idea is to keep unit mass from being crowded into cheap covering pieces. For a piece $U$ of positive diameter $r$ meeting $E$, choose $x\in U\cap E$. A ball of radius $r+\epsilon$ contains $U$; letting $\epsilon \downarrow 0$ gives ${\mu }^{\ast }(U)\le C{r}^s$, where ${\mu }^{\ast }$ is the associated outer measure. Zero-diameter pieces have zero mass by the same estimate. For any sufficiently small cover,Thus every cover costs at least $1/C$. This is the lower-bound counterpart to the explicit covers in scale-restricted Hausdorff content. The Cantor distribution supplies the needed mass in Hausdorff dimension of the Cantor set. ^[ref].

A ring of sets is closed under finite unions and differences. A premeasure on it is a nonnegative set function with value zero on the empty set that is countably additive whenever a disjoint union remains in the ring. Caratheodory extension theorem extends a premeasure to a measure on the generated sigma-algebra; this extension is unique there when the premeasure is sigma-finite. It is obtained from the outer measurewhose measurable sets satisfy ${\mu }^{\ast }(E)={\mu }^{\ast }(E\cap B)+{\mu }^{\ast }(E\setminus B)$ for every $E$. The resulting complete measurable sigma-algebra contains the sigma-algebra generated by the ring.

For the required continuity, let $B_n\downarrow \varnothing$ in the bounded-interval ring and fix $ϵ>0$. From each component $(a,b]$ of $B_n$, remove a sufficiently short initial piece and retain the compact interval $[a+\delta ,b]$. The union is a compact set $K_n\subset B_n$, and $B_n\setminus K_n$ is covered by a ring set $C_n$ of total length less than $ϵ2^{-n}$; one can take unions of $(a,a+\delta ]$. We do not incorrectly assume the compact sets themselves belong to the half-open interval ring.

Since ${\bigcap }_n{K}_n=\varnothing$ and $K_1$ is compact, the finite intersection property supplies $N$ with ${\bigcap }_{n=1}^N{K}_n=\varnothing$. For $x\in B_N$, membership in all earlier $B_n$ means some failure of membership in $K_n$ puts $x$ in $C_n$. ThusFinite additivity gives the finite subadditivity used here. Monotonicity now proves $\mu (B_n)\to 0$.

If a disjoint union $A={\bigcup }_j{A}_j$ lies in the ring, its remainders $A\setminus {\bigcup }_{j\le n}{A}_j$ decrease to the empty set. The proved continuity and finite additivity imply countable additivity, so the length function is a premeasure. The ring generates the Borel sigma-algebra and covers the line by bounded intervals of finite measure. Its unique sigma-finite extension is Borel length measure; its completion is Lebesgue measure.

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 $y\in B(c,r)$,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,Disjointness givesThe empty compact set is covered by the empty selection.

The spherical derivative of a measure at $x$ is the limitwhen 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 $\eta$, putThis 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 giveBy inner regularity of Lebesgue measure, this proves the uncentered maximal weak-type inequality

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 $\epsilon >0$ choose compact $L\subset N$ with $\mu ({\mathbb{R}}^d\setminus L)<\epsilon$. Define the remainder measure ${\eta }_{\epsilon }(A)=\mu (A\setminus L)$.

For $x\notin L$, distance from $x$ to this compact set is positive. Thus sufficiently small centered balls avoid $L$, andWrite ${\overline{D}}_s\mu (x)$ for the upper limit of the ratios. If it exceeds $\alpha$, then either $x\in L$ or $M_u{\eta }_{\epsilon }(x)>\alpha$. HenceThe right side has Lebesgue measure at most $3^d\epsilon /\alpha$, 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 $\epsilon \downarrow 0$, 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 andThis 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 $h\ne 0$, 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 $h<0$. ThusAt 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 the ordinary two-sided derivative exists andThis 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.