Hausdorff Measure and Hausdorff Dimension — Examples

There are $2^m$ basic intervals of diameter $3^{-m}$, and $3^s=2$. Thus their total cost is $2^m{3}^{-ms}=1$. The sharp Cantor theorem provides the matching lower bound, so the infimum cannot fall below one in the limit.

Since $0<s<1$ and the critical measure equals the finite value one, exponent comparison gives ${\mathcal{H}}^1(C)=0$. At level zero the cover is $[0,1]$ and also costs one; shrinking scales require arbitrarily large levels.

The construction intervals at a fixed level are disjoint, and induction in the defining recursion gives $2^n{\mathrm{\ell }}_n=\frac{1}{2}+2^{-n-1}$. Thus ${\lambda }_1(S_n)=\frac{1}{2}+2^{-n-1}$. The stages are closed, and ${\lambda }_1(S_0)=1<\mathrm{\infty }$. Their intersection is the closed set $S$.

Continuity from above yields ${\lambda }_1(S)={\lim }_n{\lambda }_1(S_n)=1\mathrm{/}2$. The equality with ${\mathcal{H}}^1$ and the finite-positive criterion give the stated measure and dimension. Thus the earlier cover lower bound has been matched by an exact measure calculation here.

If $\mathrm{\ell }=\mathrm{\mid }q-p\mathrm{\mid }>0$, the map $u\mapsto p+u(q-p)\mathrm{/}\mathrm{\ell }$ is an isometry from $[0,\mathrm{\ell }]$ onto $[p,q]$, since the distance between its images is $\mathrm{\mid }u-v\mathrm{\mid }$. Hence the segment has ${\mathcal{H}}^1$ equal to the interval length $\mathrm{\ell }$.

For $\mathrm{\ell }>0$ this value is finite and positive, so the dimension is one. If $\mathrm{\ell }=0$, the segment is the singleton $\{p\}$: its own singleton cover costs zero at exponent one and it is countable, giving dimension zero. Both closed endpoints are present in the parametrisation.

The graph map $g(x)=(x,f(x))$ satisfies $\mathrm{\mid }g(x)-g(y){\mathrm{\mid }}^2=\mathrm{\mid }x-y{\mathrm{\mid }}^2+\mathrm{\mid }f(x)-f(y){\mathrm{\mid }}^2\le (1+L^2)\mathrm{\mid }x-y{\mathrm{\mid }}^2$. Its Lipschitz constant is at most $\sqrt{1+L^2}$, which is positive even for $L=0$. Since the unit interval has Lebesgue length one, [F2] gives ${\mathcal{H}}^1([0,1])=1$ and hence the upper measure bound.

The coordinate projection $\pi :\mathrm{\Gamma }\to [0,1]$ is $1$-Lipschitz and onto. Therefore $1={\mathcal{H}}^1([0,1])\le {\mathcal{H}}^1(\mathrm{\Gamma })$. Both bounds show finite positive measure, and thus dimension one. For $L=0$ both measure bounds equal one.

As a subset of the countable rationals, $D$ is at most countable; hence ${\dim }_{H}D=0$. The endpoints zero and one are included.

Every relative neighbourhood in $[0,1]$ contains a rational point of $[0,1]$, by density (and the endpoints themselves at the ends). Thus $\bar{D}=[0,1]$, whose Lebesgue measure is one and whose dimension is consequently one.

Every length-$n$ word gives a lower-left corner $p={\sum }_{j\le n}{2}^{-j}{d}_j$ and a containing closed square $Q=p+[0,2^{-n}{]}^2$. There are $3^n$ words, giving distinct grid squares because each coordinate prefix has a unique length-$n$ binary code. Their diameters are $\sqrt{2}{2}^{-n}$; hence their $s$-cost is $3^n(\sqrt{2}{2}^{-n}{)}^s=2^{s\mathrm{/}2}$. The series converge coordinatewise by geometric tails, and these covers at arbitrarily small scales give ${\mathcal{H}}^s(K)\le 2^{s\mathrm{/}2}$.

For $u\in [0,1)$ define $e_j(u)=\lfloor 3^{j}u\rfloor -3\lfloor 3^{j-1}u\rfloor \in \{0,1,2\}$, identify these three values with the listed members of $D$, and put $T(u)={\sum }_{j\ge 1}{2}^{-j}d(e_j(u))$. Each coordinate is a limit of Borel step functions, hence Borel measurable. The vector map is Borel since preimages of open rational rectangles are Borel and those rectangles form a countable basis. Define $\mu (B)={\lambda }_1(T^{-1}(B))$ for Borel $B\subseteq {\mathbb{R}}^2$, with preimages taken in $[0,1)$. Disjoint Borel preimages prove countable additivity; thus $\mu$ is a Borel probability.

Every ternary prefix event is a half-open interval of length $3^{-n}$, hence has probability $3^{-n}$. Its image lies in the corresponding square. Also $T([0,1))\subseteq K$, so every Borel superset of $K$ has $\mu$-measure one and ${\mu }^{\ast }(K)=1$; no measurability claim about an arbitrary image is needed.

For nonempty $U$ of diameter $r$ with $2^{-n}\le r<2^{1-n}$, each coordinate projection lies in an interval of length at most $r<2\cdot 2^{-n}$. Such an interval meets at most four closed grid intervals of side $2^{-n}$, allowing all boundary contacts. Thus $U$ meets at most sixteen level-$n$ grid squares. Let $R$ be the closed coordinate bounding rectangle of $U$; its coordinate side lengths are at most $r$, so it meets at most sixteen squares. Every $u$ with $T(u)\in R$ has its own prefix square meeting $R$, so ${\mu }^{\ast }(U)\le \mu (R)\le 163^{-n}\le 16r^s$.

For a singleton use its coordinate point rectangle at arbitrarily fine levels; at most four squares contain the point, so its mass is at most $43^{-n}\to 0$. Empty sets have zero mass. The diameter estimate therefore holds also at zero. Apply mass distribution with constant sixteen and outer mass one to obtain ${\mathcal{H}}^s(K)\ge 1\mathrm{/}16$. Combined with the finite upper bound, this gives ${\dim }_{H}K=s$.

For the nonsquare positions, $a_S(n)=n-\lfloor \sqrt{n}\rfloor$. Therefore $a_S(n)\mathrm{/}n\to 1$ and the dimension formula gives ${\dim }_H{A}_S=1$, with compactness supplied by the same theorem.

The forbidden positions include every positive square and are infinite. Thus ${\lambda }_1(A_S)={\mathcal{H}}^1(A_S)=0$. The set contains zero and is a nonempty witness refuting the implication.

Here $a_S(n)=\lfloor \sqrt{n}\rfloor$, so $a_S(n)\mathrm{/}n\to 0$. The digit theorem gives compactness and dimension zero. Both the square positions and the nonsquare positions are infinite, so its uncountability conclusion applies.

Every $t>0$ lies strictly above this dimension. Hence ${\mathcal{H}}^t(A_S)=0$ for every such finite exponent. The conclusion concerns positive exponents only; at exponent zero the uncountable set is not null.

Given $y\in [0,1]$, surjectivity provides $x\in [0,1]$ with $c(x)=y$. If $x\in C$ this already suffices. Otherwise $x$ lies in a gap $(u,v)$ whose endpoints are in $C$ and $c(u)=c(x)=y$. Thus $c(C)=[0,1]$. Restricting the continuous function to $C$ preserves continuity.

The domain has dimension $\log 2\mathrm{/}\log 3<1$, whereas the image interval has positive length and dimension one. This is the claimed strict increase; no injectivity is asserted for this example.

By the Vitali theorem there is a nonmeasurable $V$. Therefore some $T\subseteq \mathbb{R}$ fails its Carathéodory splitting identity. Let $A=T\cap V$ and $B=T\setminus V$; these are disjoint and their union is $T$. Thus Lebesgue outer measure fails finite additivity on this pair. Full Choice here supplies in particular the Countable Choice hypothesis of the line comparison.

The line equality transfers this failure to ${\mathcal{H}}^1$. Add empty sets after $A,B$ to make a disjoint sequence; its sum is still ${\mathcal{H}}^1(A)+{\mathcal{H}}^1(B)$ and differs from ${\mathcal{H}}^1(T)$. Neither piece can be empty, since such a splitting would be automatic. Hence countable additivity on all subsets is false.

Let $A=[0,1]$ and $B=[2,3]$. They are disjoint compact intervals, each with positive length, so ${\dim }_{H}A={\dim }_{H}B=1$.

Countable stability applied to these two sets and empty remaining terms gives ${\dim }_H(A\cup B)=\max (1,1)=1$. This differs from $1+1=2$, refuting the asserted sum rule.

The nonsquare-position digit set has dimension one but critical measure zero. Hence dimension alone does not force positive critical measure.

The real line also has dimension one but critical measure infinity. Hence dimension alone does not force finite critical measure either. The two witnesses refute the two strict inequalities separately.

Nonnegativity, symmetry and separation for $\rho$ follow from those for $d$. For the triangle inequality, $\sqrt{a+b}\le \sqrt{a}+\sqrt{b}$ for $a,b\ge 0$, because squaring the right side gives $a+b+2\sqrt{ab}\ge a+b$. Apply this to the triangle inequality for $d$. Also $B_{\rho }(x,r)=B_d(x,r^2)$ for every $r>0$, so both metrics have identical open sets and the identity is a homeomorphism. The usual compact interval is therefore compact in both metrics.

For every subset $U\subseteq I$, ${\mathrm{diam}}_{\rho }U=({\mathrm{diam}}_{d}U{)}^{1\mathrm{/}2}$; for nonempty sets this follows from monotonicity and continuity of the square root applied to the supremum of distances, and for the empty set both sides are zero. Thus for $t\ge 0$ and $\delta >0$, the same cover families give ${\mathcal{H}}_{\rho ,\delta }^t(I)={\mathcal{H}}_{d,{\delta }^2}^{t\mathrm{/}2}(I)$. Nonempty singleton costs match also when $t=0$. Passing to the small-scale suprema yields ${\mathcal{H}}_{\rho }^t(I)={\mathcal{H}}_d^{t\mathrm{/}2}(I)$. Covers in the line may be intersected with $I$ without increasing their costs, and covers in $I$ are line covers, so the usual ambient and subspace values agree.

The unit interval has Lebesgue length one. At $t=2$ the preceding identity gives ${\mathcal{H}}_{\rho }^2(I)={\mathcal{H}}_d^1(I)=1$. The finite-positive criterion gives dimensions two and one in the two metrics. The identity is bijective and hence injective, so this is a counterexample to the asserted invariance.

Let $S=\{k^2:k\ge 1\}$ and use the set $A_S$ of the cited counterexample. It has Hausdorff dimension zero.

The same set is uncountable. Thus it satisfies the hypothesis but not the conclusion of the asserted implication.

Take the square-position binary digit set $A_S$. The cited result gives ${\mathcal{H}}^s(A_S)=0$ simultaneously for every finite $s>0$, as required by the antecedent.

That result also establishes that $A_S$ is uncountable, refuting the conclusion. The quantifier excludes exponent zero, so there is no assertion that its counting measure vanishes.