Hausdorff Measure and Hausdorff Dimension

Every cover of $B$ covers $A$, and every $\eta$-cover is a $\delta$-cover. Infima over the larger families are smaller, including when a family is empty.

The dyadic values form a nondecreasing nonnegative sequence. Its supremum $M$ exists; if finite, the definition of supremum gives eventual values above $M-\epsilon$, and if infinite, eventual values above every real bound. Thus the sequence has extended limit $M$.

For each $\delta >0$ there is $k$ with $2^{-k}\le \delta$, so ${\mathcal{H}}_{\delta }^s(A)\le M$. Conversely every dyadic value occurs among the scale values. Their suprema agree. For $A=\mathrm{\varnothing }$ all values are zero; no part divided by $s$, so $s=0$ is included.

The empty cover has cost zero at every scale; a cover of a larger set covers each subset. Hence both functions vanish at the empty set and are monotone. This uses no positive-exponent assumption.

Fix $\delta$ and $(A_j)$. If ${\sum }_j{\mathcal{H}}_{\delta }^s(A_j)=\mathrm{\infty }$, subadditivity is automatic. Otherwise, for $\epsilon >0$ select for each $j$ a cover with cost at most ${\mathcal{H}}_{\delta }^s(A_j)+\epsilon 2^{-j-1}$; select an enumerated cover, so the resulting double family is countable. Empty $A_j$ may use the empty family.

Flatten these covers along an enumeration of the pairs of indices. Every finite subfamily cost is bounded by the corresponding iterated sum, and every finite rectangle is eventually included; thus the nonnegative sums agree. The union has scale cost at most ${\sum }_j{\mathcal{H}}_{\delta }^s(A_j)+\epsilon$. Letting $\epsilon$ decrease to zero proves the fixed-scale assertion.

For finite $\delta$, the same inequality is at most ${\sum }_j{\mathcal{H}}^s(A_j)$. This bound is independent of $\delta$; taking the supremum proves the claimed inequality, including infinite right sides. Together with the first step these are all the outer-measure axioms.

Let $h=d(A,B)>0$. A covering set of diameter at most $\delta <h$ cannot meet both $A$ and $B$. Partition any cover of their union by which set it meets, discarding sets meeting neither. Its cost is at least ${\mathcal{H}}_{\delta }^s(A)+{\mathcal{H}}_{\delta }^s(B)$; if no cover exists the inequality still holds.

Take small-scale suprema in that inequality. The supremum of the sums of the two nondecreasing scale values is the sum of their suprema: approximate both finite lower bounds at one common scale; this also proves the assertion when one supremum is infinite. Subadditivity gives the opposite inequality. Empty sets use the outer-measure zero axiom. Thus the metric condition holds, also at $s=0$.

The metric criterion gives Borel measurability, and the Carathéodory theorem gives completeness on the full measurable domain.

For a finite $A$ of size $N$, its singleton cover costs $N$ at every scale, including $N=0$. If $N\ge 2$, choose $\delta$ smaller than the minimum distance between distinct points; every cover then has at least $N$ members. For $N=1$ any cover needs one member. Thus ${\mathcal{H}}^0(A)=N$.

If $A$ is infinite, for each positive integer $N$ it contains an $N$-point subset. The preceding lower bound gives ${\mathcal{H}}^0(A)\ge N$ for every $N$, hence infinity. The formula is exactly the counting set function.

For any test set $T$ and subset $E$, if $T$ is finite its partition into $T\cap E$ and $T\setminus E$ splits its cardinality. If $T$ is infinite at least one piece is infinite, and both sides of the splitting identity are infinity. This is the Carathéodory criterion for every $E$.

If ${\mathcal{H}}^s(A)=\mathrm{\infty }$, take $G=X$. If $A=\mathrm{\varnothing }$, take $G=\mathrm{\varnothing }$. If $s=0$ and the measure is finite, $A$ is finite, hence closed and Borel. In Euclidean space a finite set is a $G_{\delta }$, by intersecting its open $1\mathrm{/}k$-neighbourhoods.

In the remaining case $s>0$ and $M={\mathcal{H}}^s(A)<\mathrm{\infty }$, choose for each $k\ge 1$ a $2^{-k}$-cover $(U_{kj})$ of cost at most $M+2^{-k}$. Replacing each set by its closure preserves diameter: approximate two closure points by original points and use the triangle inequality. Set $G={\bigcap }_k{\bigcup }_j\overline{U_{kj}}$. This is Borel and contains $A$.

For fixed $\delta >0$ and all sufficiently large $k$, the $k$th closed cover also covers $G$ at scale $\delta$. Hence ${\mathcal{H}}_{\delta }^s(G)\le M+2^{-k}$, so ${\mathcal{H}}_{\delta }^s(G)\le M$. Take the supremum over $\delta$ and use monotonicity to get equality. Infimising Borel-superset measures gives the displayed identity: every such value is at least ${\mathcal{H}}^s(A)$ and this $G$ attains it.

For Euclidean $X$ in the finite positive-exponent case, enlarge $U_{kj}$ to the open neighbourhood $V_{kj}=\{x:d(x,U_{kj})<{\eta }_{kj}\}$ with $0<{\eta }_{kj}<2^{-k-2}$ and $(\mathrm{diam}{U}_{kj}+2{\eta }_{kj}{)}^s\le (\mathrm{diam}{U}_{kj}{)}^s+2^{-k-j-1}$. Continuity of the positive power at every nonnegative finite base supplies these choices, including singleton sets. Then $G={\bigcap }_k{\bigcup }_j{V}_{kj}$ is $G_{\delta }$, each covering diameter is at most $2^{1-k}$, and each cost is at most $M+2^{1-k}$. The same fixed-scale argument proves equality.

First suppose ${\mathcal{H}}^s(E)<\mathrm{\infty }$. Choose a Borel hull $B$ of $E$. Measurability of $E$ gives ${\mathcal{H}}^s(B)={\mathcal{H}}^s(E)+{\mathcal{H}}^s(B\setminus E)$, hence $B\setminus E$ is null. Choose a null Borel hull $N$ of that difference; then $B\setminus N\subseteq E\subseteq B$ is the required sandwich. This also covers empty or null $E$.

For the Euclidean assertion take $G={\bigcap }_{i\ge 1}{O}_i\supseteq E$, with $O_i$ open and ${\mathcal{H}}^s(G\setminus E)=0$. Each $O_i$ has an increasing closed exhaustion: if its complement is nonempty, use $F_{ij}=\{x:\mathrm{\mid }x\mathrm{\mid }\le j,d(x,O_i^c)\ge 1\mathrm{/}j\}$; for $O_i={\mathbb{R}}^n$ use the closed balls. Continuity of the distance function follows from its triangle-inequality Lipschitz bound. These sets exhaust $O_i$.

For sigma-finite $E$, write $E={\bigcup }_j{E}_j$ with measurable finite-measure $E_j$ and apply the first step to each. The unions of the lower and upper hulls are Borel, and their difference is contained in the union of the null differences. Thus it is null. Only the finite pieces required subtraction; this includes infinite $E$.

Fix $\epsilon >0$. Continuity from below on $E\cap F_{ij}$, with ${\mathcal{H}}^s(E)<\mathrm{\infty }$, gives $j_i$ such that ${\mathcal{H}}^s(E\setminus F_{i{j}_i})<\epsilon 2^{-i}$. The closed set $F={\bigcap }_i{F}_{i{j}_i}$ lies in $G$ and loses less than $\epsilon$ of $E$. Its excess $F\setminus E$ is null. Take a null $G_{\delta }$ hull $N$ of that excess. Then $F\setminus N$ is $F_{\sigma }$, lies in $E$, and loses less than $\epsilon$ of $E$.

Apply this for $\epsilon =1\mathrm{/}k$ and take the union of the resulting $F_{\sigma }$ sets; it is an $F_{\sigma }$ subset $H$ of $E$ with ${\mathcal{H}}^s(E\setminus H)=0$. Express $H$ as an increasing union of closed sets (replace any closed sequence by its finite unions). Continuity from below now gives a closed subset of $E$ with deficit less than any prescribed positive number. The argument applies to $s=0$ as well; in that case finite-measure $E$ is finite.

Suppose $s>0$ and ${\mathcal{H}}_{\mathrm{\infty }}^s(A)=0$. Given finite $\delta >0$ and $\epsilon >0$, take a cover with cost less than $\min (\epsilon ,{\delta }^s)$. Each diameter is strictly below $\delta$, so ${\mathcal{H}}_{\delta }^s(A)<\epsilon$. Therefore every scale value is zero.

When $s=0$, any nonempty member costs one. Content less than one forces the empty family, hence $A=\mathrm{\varnothing }$. All its scale values are zero. This deals with empty and singleton possibilities without division by the exponent.

If all finite-scale values vanish their supremum vanishes. Conversely, a zero supremum forces all those nonnegative values, and then the smaller unrestricted content, to vanish. These implications complete both equivalences.

For $L>0$, replace every member $U$ of a $\delta$-cover of $A$ by $f(U\cap D)$ and discard empty images. The image diameter is at most $L\mathrm{diam}U$. For positive $s$ its cost is at most $L^s$ times the old cost; for $s=0$ each retained member still costs one.

Infimising gives ${\mathcal{H}}_{L\delta }^s(f(A))\le L^s{\mathcal{H}}_{\delta }^s(A)$; if the right scale infimum is infinite the inequality is automatic. Let $\delta$ tend to zero to obtain the assertion, since $L$ is finite and positive.

For $L=0$, any two image points have distance zero, so a nonempty image is a singleton. Its singleton cover costs zero for $s>0$. For $s=0$, image cardinality cannot exceed domain cardinality, whether finite or infinite. Empty images have zero measure at every exponent.

Since $c>0$, equality of images implies distance zero in $D$, so $f$ is injective. Its inverse on $f(D)$ is $1\mathrm{/}c$-Lipschitz. Apply the inequality to $f$ and its inverse to obtain both bounds and hence equality; all scalar multipliers are positive, so infinite values cause no indeterminate product.

An ambient cover can be intersected with the subspace and its empty members discarded, without increasing cost or diameter. Conversely every subspace cover is an ambient cover. Infima at each scale, and then their suprema, agree. The empty set and $s=0$ obey the same comparison. Taking $c=1$ gives isometries and translations; a Euclidean dilation by $c>0$ scales distances by $c$.

For $0<r\le \delta$, $r^t=r^s{r}^{t-s}\le r^s{\delta }^{t-s}$. At $r=0$ the left side is zero; the right is zero for $s>0$ and ${\delta }^t$ for $s=0$. Thus the inequality holds for every admissible covering member. Summation and infimisation prove the fixed-scale bound; if no cover exists its right side is infinity.

If $M={\mathcal{H}}^s(A)<\mathrm{\infty }$, then ${\mathcal{H}}_{\delta }^t(A)\le {\delta }^{t-s}M$. Given any fixed $\eta >0$, use all $0<\delta \le \eta$ and monotonicity to bound ${\mathcal{H}}_{\eta }^t(A)$ by a quantity tending to zero. Every scale value is therefore zero and so is the supremum. This includes $M=0$ and the empty set.

If $s<d$ and ${\mathcal{H}}^s(A)$ were finite, choose a finite $t$ strictly between $s$ and $d$ (also possible for $d=\mathrm{\infty }$). Then ${\mathcal{H}}^t(A)=0$, contrary to $d$ being the infimum of the zero exponents. Hence ${\mathcal{H}}^s(A)=\mathrm{\infty }$.

If $d<s<\mathrm{\infty }$, the zero-exponent set is nonempty and contains $u<s$ by its infimum property. Exponent comparison gives ${\mathcal{H}}^s(A)=0$. Thus when $d=0$ all positive exponents vanish, and when $d=\mathrm{\infty }$ every finite exponent has infinite measure.

The first two steps place every finite-measure exponent at least $d$, and every exponent strictly greater than finite $d$ among the finite-measure exponents. Their infimum is $d$, also when the set is empty. Similarly the infinite-measure exponents lie at most $d$ and contain every nonnegative exponent strictly below $d$. Their supremum is $d$; for $d=0$ it is zero whether that set is empty or consists of zero. Finite positive measure at $s$ excludes $s<d$ and $s>d$, hence forces equality. Empty $A$ has $d=0$ and no infinite-measure exponent.

If $A\subseteq B$, every exponent with ${\mathcal{H}}^s(B)=0$ also has ${\mathcal{H}}^s(A)=0$. Infimising zero exponents gives ${\dim }_{H}A\le {\dim }_{H}B$, including empty zero-exponent sets. Thus the dimension of the union is at least $D={\sup }_k{\dim }_H{A}_k$.

If $D=\mathrm{\infty }$ that lower bound is equality. If $D<\mathrm{\infty }$, for every $t>D$ all ${\mathcal{H}}^t(A_k)$ vanish, so countable subadditivity makes their union null. Its dimension is at most every $t>D$, hence at most $D$. This includes $D=0$.

An at most countable set has a cover by its singletons, whose total cost is zero for each $t>0$. The empty set has the empty cover. Thus their dimensions are zero; one point and every finite set are included.

For a positive Lipschitz constant and finite $d={\dim }_{H}A$, every $t>d$ has ${\mathcal{H}}^t(f(A))=0$, so ${\dim }_{H}f(A)\le d$. For $d=\mathrm{\infty }$ the inequality is automatic. A constant map has empty or singleton image, with dimension zero from its singleton covers, so its case also holds.

The two-sided bounds make $f$ Lipschitz with constant $b$ and its inverse on $f(D)$ Lipschitz with constant $1\mathrm{/}a$. Apply the first step in both directions. Empty sets and the value zero are allowed throughout.

If a nonempty bounded $U$ has diameter $r$, its $i$th coordinate ranges between an infimum $a_i$ and supremum $b_i$ with $b_i-a_i\le r$. Hence $U\subseteq {\prod }_i[a_i,b_i]$ and ${\lambda }_n^{\ast }(U)\le {\prod }_i(b_i-a_i)\le r^n$. If $r=0$ this is a zero-volume singleton box.

For every finite-scale cover of $A$, countable subadditivity gives ${\lambda }_n^{\ast }(A)\le {\sum }_j(\mathrm{diam}{U}_j{)}^n$. Infimising and then taking the small-scale supremum gives the lower bound; absent covers give the same inequality with infinity on the right. The empty set has both values zero.

Partition $(0,1{]}^n$ into $m^n$ half-open cubes of side $1\mathrm{/}m$. Each has diameter $\sqrt{n}\mathrm{/}m$ (the supremum of corner distances), so the total cost is $m^n(\sqrt{n}\mathrm{/}m{)}^n=n^{n\mathrm{/}2}$. Choose $m$ large for any prescribed positive scale. Thus the upper bound holds, while the lower bound is the unit box volume one. For $n=1$ both bounds equal one.

Only the reverse inequality needs proof. If ${\lambda }_1^{\ast }(A)=\mathrm{\infty }$, the lower inequality already gives equality. Otherwise choose an elementary-set cover of total volume less than ${\lambda }_1^{\ast }(A)+\epsilon$. Each finite-volume elementary set is a finite disjoint union of bounded half-open intervals; zero-volume empty pieces can be discarded. This is the elementary algebra and its volume appearing in the defining infimum.

Fix $\delta >0$. Subdivide each such interval into finitely many half-open intervals of lengths at most $\delta$, without changing the sum of lengths. Flatten the resulting countable family. It is an admissible Hausdorff cover of cost at most ${\lambda }_1^{\ast }(A)+\epsilon$. Hence ${\mathcal{H}}_{\delta }^1(A)\le {\lambda }_1^{\ast }(A)$ after letting $\epsilon$ decrease to zero. Taking the supremum over $\delta$ proves equality. For empty $A$ use the empty cover.

For every test set $T$ the two outer values in the Carathéodory splitting identity are identical. Thus a set satisfies that identity for one outer measure if and only if it does for the other, and the restricted values agree. Both restrictions are their complete Carathéodory measures.

Set $c_n={\mathcal{H}}^n((0,1{]}^n)$, which is finite and positive. The Borel function $\nu (B)=c_n^{-1}{\mathcal{H}}^n(B)$ is a measure, since multiplication by a fixed positive scalar preserves nonnegative sums.

Translations preserve $\nu$ and $\nu ((0,1{]}^n)=1$. The uniqueness theorem therefore gives $\nu (B)={\lambda }_n(B)$ for every Borel $B$. Its hypotheses include Countable Choice, the Borel domain, and precisely the half-open normalising cube used here.

For arbitrary $A$, take its Hausdorff Borel hull $G$. Then $c_n{\lambda }_n^{\ast }(A)\le c_n{\lambda }_n(G)={\mathcal{H}}^n(A)$. Take instead a Lebesgue Borel hull $H$; then ${\mathcal{H}}^n(A)\le {\mathcal{H}}^n(H)=c_n{\lambda }_n^{\ast }(A)$. Both inequalities remain valid for infinite values, and the empty set has zero value.

The line equality gives $c_1=1$ by evaluation on $(0,1]$. The higher-dimensional proof used only the finite positive cube bounds, so it has established no sharper constant.

Every nondegenerate bounded cube has finite positive Lebesgue volume, hence finite positive ${\mathcal{H}}^n$ and dimension $n$. Countably many such cubes cover ${\mathbb{R}}^n$, so its dimension is $n$ by countable stability.

Monotonicity bounds every subset above by $n$, including the empty set. If its Lebesgue outer measure is positive, its ${\mathcal{H}}^n$ is positive, possibly infinite, and the critical-exponent theorem bounds its dimension below by $n$. Thus equality holds. The statements include $n=1$.

The function ${\mu }^{\ast }$ vanishes on the empty set, is monotone, and agrees with $\mu$ on Borel sets: inclusion gives the lower inequality, while the set itself is a candidate hull. It is countably subadditive: for finite ${\sum }_j{\mu }^{\ast }(E_j)$ choose Borel supersets $B_j$ of costs at most ${\mu }^{\ast }(E_j)+\epsilon 2^{-j-1}$ and use $\mu ({\bigcup }_j{B}_j)\le {\sum }_j\mu (B_j)$. The latter follows by disjointifying the Borel sets and applying countable additivity. Infinite right sides are automatic. Let $\epsilon$ decrease to zero. Thus it is an outer measure.

Fix $0<\delta <r_0$ and any admissible cover $(U_j)$ of $A$. Subadditivity and the assumed bound give ${\mu }^{\ast }(A)\le {\sum }_j{\mu }^{\ast }(U_j)\le C{\sum }_j(\mathrm{diam}{U}_j{)}^s$. Infimising gives ${\mathcal{H}}_{\delta }^s(A)\ge {\mu }^{\ast }(A)\mathrm{/}C$, even if no cover exists. Taking the supremum and using the critical-exponent criterion proves both conclusions; at $s=0$ the dimension bound is the automatic nonnegativity.

For the ball hypothesis, fix a nonempty $U$ of diameter $d<r_0$ and one $x\in U$. For every $d<r<r_0$, $U\subseteq B(x,r)$, so ${\mu }^{\ast }(U)\le C{r}^s$. Let $r$ decrease to $d$. If $d=0$ and $s>0$ the bound is zero; if $s=0$ it is $C$, as required by the covering-cost convention. This proves the sufficient ball condition without evaluating $\mu$ on a non-Borel set.

The left endpoint has digits $2b_1,\dots ,2b_m$ followed by zeros; the right endpoint has that same prefix followed by twos, since ${\sum }_{j>m}23^{-j}=3^{-m}$. Their Cantor-function values are respectively $q={\sum }_{j=1}^m{b}_j{2}^{-j}$ and $q+{\sum }_{j>m}{2}^{-j}=q+2^{-m}$.

The distribution-function interval formula and absence of endpoint atoms give ${\mu }_c(I_b)=(q+2^{-m})-q=2^{-m}$. This also applies at the endpoints zero and one of the extended distribution function; for $m=0$ it agrees with probability mass one.

Here $0<s<1$ and $3^s=2$. For $a,b,c\ge 0$ with $a,c\le b$, one has $a^s+c^s\le (a+b+c{)}^s$. Indeed the function $(x+y{)}^s-x^s$ is nonincreasing in $x>0$ for fixed $y\ge 0$, by its derivative; continuity extends this to $x=0$. Increasing $a$ to $b$ and then $c$ to $b$ can only decrease $(a+b+c{)}^s-a^s-c^s$, whose final value is $b^s(3^s-2)=0$. If $b=0$ all three numbers are zero.

Fix $m$. For each basic interval $J$ at a level $\mathrm{\ell }\le m$ and each interval $I$, let $N_J(I)$ count the level-$m$ basic descendants contained in $I\cap J$. We prove $2^{-m}{N}_J(I)\le \mathrm{diam}(I\cap J{)}^s$ by induction upwards from level $m$. At that level the count is zero or one; a count of one forces diameter at least $3^{-m}$ and hence the desired bound. Empty intersections have count and diameter zero.