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The law of ${\sum }_{j\ge 1}2{\omega }_j{3}^{-j}$, for independent fair binary digits ${\omega }_j$, is supported on the Cantor set. It is invariant and ergodic under $T_3$, as the image of a Bernoulli shift. Each permitted ternary cylinder of length $n$ has measure $2^{-n}$, giving entropy rate $\log 2$. The Host equidistribution theorem makes almost every point a normal number in base $2$, while its ternary digits omit $1$.

A Cantor cylinder fixes finitely many of the zero-or-two ternary digits in the standard Cantor set. These sets form finite clopen partitions of diameter at most $3^{-n}$ at level $n$, and a base for the relative topology. They permit uniform approximation of continuous functions by locally constant functions.

The Cantor distribution is the law of a random variablewhere the $B_n$ are independent Bernoulli random variables taking values zero and one with equal probability. Its values lie in the Cantor set, and its cumulative distribution function is the Cantor function. Every individual value has probability zero, but the Cantor set has probability one despite having zero Lebesgue measure. A probability density function on that set would integrate to zero, so it cannot describe this law. The example motivates the distinction between a continuous probability distribution and an absolutely continuous probability distribution.

Cantor space is the binary sequence space with the product topology of discrete two-point factors. Here the Bernoulli-space synonym refers to this topology, without selecting a particular probability measure. The metric $d(x,y)={\sum }_{j\ge 1}{2}^{-j}|x_j-y_j|$ induces the topology. A diagonal subsequence argument gives compactness. Fixing a finite prefix gives a clopen cylinder set. The map $x\mapsto {\sum }_{j}2x_j{3}^{-j}$ is a homeomorphism to the usual Cantor set.

For the middle-third Cantor set $C$ and positive integer $d$, $C^d$ is compact and totally disconnected. A connected subset projects to a point in each coordinate and is therefore itself a point. At level $n$, the product has $2^{dn}$ cubes of side $3^{-n}$. These covers give dimension at most $d\log 2/\log 3$. The product digit probability assigns each cube mass $2^{-dn}$; a ball of radius comparable to $3^{-n}$ meets only a bounded number of cubes, so its mass is at most a constant times $r^{d\log 2/\log 3}$. The cover-cost argument for Hausdorff dimension proves the matching lower bound. Thus compact totally disconnected Euclidean subsets can have arbitrarily large Hausdorff dimension when the ambient dimension is allowed to increase.

Every real continuous function of supremum norm at most one on the Cantor set is a uniform limit of convex combinations of continuous sign functions. Approximate on a finite Cantor cylinder partition by values $a_j\in [-1,1]$. The sign vector $s$ has weight ${\prod }_j(1+s_j{a}_j)/2$, whose coordinate means are $a_j$. This proves a closed convex hull identity without weak compactness.

For compact Hausdorff $K$, the extreme points of the real unit ball in the space of continuous functions on a compact space are precisely its continuous sign-valued functions. A point with $|f|<1$ permits a nonzero continuous bump perturbation supported where there is uniform slack, while pointwise equality at an endpoint of $[-1,1]$ forces both functions in a midpoint decomposition to agree. For the Cantor set, clopen indicators give the perturbations directly.

Choose a strictly decreasing positive sequence $q_0=1,q_1,q_2,\dots$. Begin with $C_0=[0,1]$ and at step $n+1$ keep two closed end intervals of length $q_{n+1}/2^{n+1}$ from every component interval of $C_n$. This constructs a compact set $C={\bigcap }_n{C}_n$ withThe second equality is continuity from above of a measure. Every stage has positive gaps and maximum component length at most $2^{-n}$, so $C$ contains no nondegenerate interval and is a nowhere dense set. It has no isolated points: both endpoints of the stage-$n$ component containing a point remain in $C$, and at least one distinct endpoint is within $2^{-n}$. Thus its topological sparseness is compatible with positive Lebesgue measure.

Taking $q_n=(2/3{)}^n$ gives the usual Cantor set of measure zero. Taking $q_n=1/2+2^{-(n+1)}$ gives a generalized construction of measure $1/2$. This is one useful family of generalized constructions; the removed intervals need not have the same proportions at every stage.

Geometric measure theory uses measures and analysis to study geometric objects that need not be smooth. Its measuring rules can distinguish a curve from its surrounding plane or capture the fine structure of a fractal. Hausdorff measure supplies size adapted to an exponent, while Hausdorff dimension identifies the exponent at which that size changes character.

The Cantor set is an elementary starting point: it has zero Lebesgue measure but positive Hausdorff dimension. In Hausdorff dimension of the Cantor set, explicit small covers give an upper bound and a distributed probability measure supplies the matching lower bound. This combination of geometric coverings and analytic control is useful even before more advanced tools for irregular surfaces are introduced.

The Hausdorff dimension is the critical exponent at which Hausdorff measure changes from infinite to zero:with the standard endpoint convention for an empty supremum. This measures fine-scale geometric size rather than the number of coordinates used to describe the set. A line segment has Hausdorff dimension one; a square has dimension two; the middle-third Cantor set has dimension $\log 2/\log 3\approx 0.63093$.

Why can a dimension be fractional? At a scale reduced by a factor of three, the Cantor set needs twice as many construction intervals. The balance $2(1/3{)}^d=1$ gives $d=\log 2/\log 3$, between the growth rates of finitely many points and a whole interval. Making this argument rigorous requires a cover upper bound and the mass distribution principle lower bound, as in Hausdorff dimension of the Cantor set.

At the critical exponent itself, Hausdorff measure can be zero, finite and positive, or infinite; dimension does not specify that endpoint value. For example, finite sets and countably infinite sets both have dimension zero, although their counting measures differ. Dimension is also different from a finite-resolution estimate: box-counting dimension is another asymptotic definition, and fitting a scaling slope over a handful of observed scales does not establish either dimension.

The metric matters. Multiplying every distance by a fixed positive constant multiplies ${\mathcal{H}}^s$ by that constant to the power $s$ and preserves Hausdorff dimension. A Lipschitz map cannot increase it, because mapped cover diameters increase by at most a fixed factor. A bi-Lipschitz map therefore preserves it by applying this fact in both directions. These properties make Hausdorff dimension useful for comparing geometric models while controlling how their measurements change.

The middle-third Cantor set has Hausdorff dimensionIts Lebesgue measure is zero, yet $0<{\mathcal{H}}^d(C)<\infty$. These assertions measure different aspects of size: interval length is too coarse to distinguish this set from a null set, while the critical fractional measuring rule captures its remaining structure.

At level $n$, the construction covers $C$ by $2^n$ intervals of diameter $3^{-n}$. Their total $s$-cost isFor $s>d$ it tends to zero, proving the upper dimension bound. At $s=d$ it equals one, giving ${\mathcal{H}}^d(C)\le 1$. For $s<d$ this particular cover's cost grows, but that alone would not prove a lower bound on Hausdorff measure.

For the matching lower bound, use the Cantor distribution, which assigns mass $2^{-n}$ to each level-$n$ Cantor cylinder. An interval of length $r$ with $3^{-(n+1)}<r\le 3^{-n}$ meets at most two level-$n$ construction intervals, since their gaps are at least $3^{-n}$. Its mass is therefore at mostEvery covering set of diameter $r$ in the line lies in an interval of length $r$, so any small cover must have total $d$-cost at least $1/4$. This is the mass distribution principle in its interval form: $1/4\le {\mathcal{H}}^d(C)\le 1$, sufficient to prove the claimed dimension. The constants are convenient bounds, rather than an assertion of the exact critical Hausdorff measure. In the diagram, the right panel is a finite-stage approximation to the Cantor function, showing how probability is shared while the middle third receives none.

The reusable pattern is to obtain an upper bound by constructing covers and a lower bound by distributing mass. The same strategy underlies Cartesian powers of the Cantor set and the dimension formula for a finite similarity iterated function system satisfying the open set condition.

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.

A Lebesgue null set has zero Lebesgue measure. Since Lebesgue measure is complete, every subset of one is also measurable and null. The Cantor set is an uncountable example with positive Hausdorff dimension; the rational numbers form a countable, dense example with Hausdorff dimension zero. Being null does not imply being empty or having finitely many points.

For $f_c(z)=z^2+c$, the Mandelbrot set is the set of parameters for which the finite critical orbit $0,f_c(0),f_c^2(0),\dots$ remains bounded. It is a compact connected parameter set. Its main cardioid consists of parameters with an attracting fixed point, with a period-two bulb attached on the left, smaller bulbs attached recursively and fine filamentary structure. It is symmetric about the real axis and meets that axis in $[-2,1/4]$. The quadratic filled Julia set is connected exactly when $c$ belongs to this parameter set; outside it, the rational critical point escapes and the Julia set is a Cantor set. Thus it organizes the contrasting phase-space dynamics of the quadratic family, rather than itself being a Julia set in the dynamical $z$-plane.

The affine map $h(z)=1/2-z$, its own inverse, givesThis value is unique even among Möbius conjugacies. The finite fixed point of $P$ is zero and its periodic-orbit multiplier is one. Holomorphic conjugacy preserves that periodic-orbit multiplier. Infinity has periodic-orbit multiplier zero for a quadratic polynomial, so a corresponding multiplier-one fixed point of $w^2+c$ must be finite; it obeys $2w=1$ and $w^2+c=w$, forcing $c=1/4$. To locate it explicitly, a quadratic fixed point of periodic-orbit multiplier $\lambda$ gives $c=\lambda /2-{\lambda }^2/4$. The boundary parametrization $|\lambda |=1$ has its cusp at $\lambda =1$, so $c=1/4$ is the right-hand cusp on the Mandelbrot boundary. It is in the set: the critical orbit starts at zero, stays in $[0,1/2]$ and increases to the fixed point $1/2$. For real $c>1/4$, the increment $f_c(x)-x=(x-1/2{)}^2+c-1/4$ is strictly positive; a bounded critical orbit would have a limit solving $x^2+c=x$, which has no real solution. These nearby real parameters escape, proving that the cusp belongs to the boundary of the full Mandelbrot set as well.

The fixed points of $P$ on the sphere are zero and infinity. There is no neighborhood of zero on which all iterates converge to zero, even pointwise: a negative real initial point satisfies $P(x)=x-x^2<x<0$, so its iterates decrease without a finite limit and escape to infinity. Every neighborhood of zero contains such points. At infinity there is a uniformly attracting neighborhood in the chordal metric. In the reciprocal coordinate $w=1/z$,For $|w|\le 1/4$, its modulus is at most $|w|/3$. Thus the reciprocal iterates tend uniformly to zero there, proving uniform convergence to infinity on the corresponding sphere neighborhood.

For the given disc, $|z-1/2|<1/2$ is equivalent to $|z{|}^2<\Re z$, and hence to $\Re (1/z)>1$. Under $w=1/z$ the iteration becomesIf $\Re w>1$, then $\Re (1/(w-1))>0$, so $\Re T(w)>\Re w+1$. Consequently the half-plane is forward invariant andThis proves uniform convergence to zero on the disc and therefore $\Delta \subset F(P)$. It is the explicit parabolic disk for z minus z squared.

Let $U=F_{\Delta }$. The iterate family is normal on the connected open set $U$. Every subsequential limit along $n_j\to \infty$ equals zero on $\Delta$; by the identity theorem it is zero throughout $U$. If the full sequence failed to converge uniformly on some compact subset, a further normal subsequence would contradict that conclusion. Hence $P^n\to 0$ locally uniformly throughout $U$. Complete invariance of the Fatou set puts $P(U)$ in a Fatou component. Since $P(\Delta )\subset \Delta$, that component is $U$, so $P(U)\subseteq U$.

We prove the reverse-image assertion, rather than assuming that a basin has only one component. Each connected component $V$ of $P^{-1}(U)$ maps properly to $U$: for compact $K\subset U$, the sphere preimage $P^{-1}(K)$ is compact, and its intersection with $V$ is closed in it because $V$ is a relatively closed component of $P^{-1}(U)$. Its image is open by holomorphic openness. To prove it is closed in $U$, lift a convergent sequence of image points to preimages in $V$. Their images eventually lie in a compact neighborhood inside $U$; properness gives a convergent preimage subsequence with limit in $V$ and the desired limiting image. Thus the image is also closed, and equals the connected target $U$. Moreover $U$ itself is one such component: it lies in $P^{-1}(U)$ by forward invariance, and any connected inverse-image component containing it lies in $F(P)$ and hence in $U$ by maximality.

Now $1/4\in \Delta$ and its entire preimage is $\{1/2\}$, counted twice. Every component $V$ of $P^{-1}(U)$ must contain a preimage of $1/4$, so every one must be $U$. ThereforeThis is the Fatou component containing an entire fiber is completely invariant argument. The convergence on this component is locally uniform: it cannot be uniform on all of $U$, because $P^n(U)=U$ for every $n$ and $U$ contains nonzero points.

Finally suppose zero belonged to the Fatou set. A normal subsequence on a disc around zero would have limit zero, since that disc meets $\Delta$ in an open set where the entire sequence tends to zero. The identity theorem forces this limit to be zero on the whole disc. Uniform convergence in a smaller finite coordinate disc would then imply derivative convergence at zero by the Cauchy integral formula. Butfor every $n$, whereas the derivative of the zero limit is zero. This contradiction proves

Using the diameter normalization, define ${\mathcal{H}}_{\delta }^d(E)=\inf \{{\sum }_j(\mathrm{diam}{U}_j{)}^d:E\subseteq {\bigcup }_j{U}_j,\mathrm{diam}{U}_j\le \delta \}$ and the Hausdorff measure ${\mathcal{H}}^d(E)={\lim }_{\delta \downarrow 0}{\mathcal{H}}_{\delta }^d(E)$. The Hausdorff dimension is $\inf \{d:{\mathcal{H}}^d(E)=0\}$, equivalently $\sup \{d:{\mathcal{H}}^d(E)=\infty \}$.

The Cantor set is the intersection of the sets left by successively removing open middle thirds from $[0,1]$. At level $k$ it has $2^k$ Cantor cylinders of diameter $3^{-k}$. For $s=\log 2/\log 3$, their total $s$-cost is $2^k{3}^{-ks}=1$, so ${\mathcal{H}}^s(C)\le 1$.

The random series $\xi ={\sum }_{n\ge 1}{X}_n{3}^{-n}$ converges absolutely, and its ternary digits are all zero or two. Thus $\xi \in C$ and its law is the Cantor Bernoulli measure. Each level-$k$ cylinder has probability $2^{-k}$. Distinct such intervals have separation at least $3^{-k}$; a set $U$ of smaller diameter meets at most one. If $3^{-(k+1)}\le \mathrm{diam}U<3^{-k}$, thenSets of diameter zero have probability zero, since the measure has no measure atoms; diameters at least one give the bound trivially. For arbitrary nonmeasurable sets use outer probability. Any countable cover of $C$ therefore satisfies $1\le {\sum }_j{\mathbb{P}}^{\ast }(\xi \in U_j)\le 2{\sum }_j(\mathrm{diam}{U}_j{)}^s$. HenceThe last implication follows by multiplying the $s$-costs by powers of the cover diameter: measures above $s$ vanish and those below $s$ are infinite.

For fractal geometry, let $C_n$ be the $n$th middle-third construction of the Cantor set. It contains $2^n$ intervals of length $3^{-n}$, with total length $(2/3{)}^n$. Each finite $C_n$ contains intervals, so its Hausdorff dimension is one. The limiting set $C={\bigcap }_n{C}_n$ has zero length andIndeed, $2^n$ intervals of scale $3^{-n}$ cover it, giving the upper dimension bound from $2^n{3}^{-ns}\to 0$ whenever $s>\log 2/\log 3$. For the lower bound, assign mass $2^{-n}$ to each level-$n$ interval. Any sufficiently small interval of length $r$ meets only a bounded number of construction intervals at the comparable scale, giving a bound $\mu (I)\le K{r}^{\log 2/\log 3}$. Covering $C$ by such intervals then forces their dimension-weighted sizes to sum to at least $1/K$, yielding the matching lower bound. The noninteger dimension is deducible from a simple iterative rule even though it is absent from every finite-stage set at arbitrarily fine resolution. A finite construction nevertheless exhibits the same scaling over intermediate resolutions. Thus the order of idealization and probing ever smaller scales matters: an actual finite-resolution pattern can have useful fractal behaviour without possessing the exact limiting dimension at all scales.

For a phase transition, finite-spin partition functions are finite sums of positive exponentials at real finite temperature and field. Their free energies are therefore analytic there. Genuine thermodynamic nonanalyticity can appear only after a suitable infinite-system limit. The Curie–Weiss model gives an explicit demonstration. With $s_i=\pm 1$, takeGrouping configurations by $m=N^{-1}{\sum }_i{s}_i$ and using their binomial multiplicities gives the limiting free energy as the minimum ofThe number of possible magnetizations grows only linearly with $N$, so their largest exponential contribution determines the limiting free energy. Differentiating the variational expression givesAt $h=0$ and $\beta J>1$, its two stable minima have magnetizations $\pm m_{\ast }\ne 0$. Every finite zero-field system instead has exactly zero mean magnetization by spin-inversion symmetry. ConsequentlyThis noncommutation of limits explains how spontaneous symmetry breaking emerges. Below the critical temperature, the limiting free energy has a cusp in $h$ at zero, while finite systems have a sharp but smooth crossover. Away from the critical point, the relative weights of the two phases behave approximately as $e^{2\beta Nh{m}_{\ast }}$; the rounded magnetization is correspondingly about $m_{\ast }\tanh (\beta Nh{m}_{\ast })$. Exact singularity belongs to the idealized limit, whereas the narrow crossover and long-lived phases can be physically useful at finite size. The theoretical reduction is the derivation and its approximation control, not a claim that a finite analytic partition function is already nonanalytic.

For superselection, consider an infinite spin lattice with quasi-local observable algebra $\mathcal{A}$. These are zero-temperature equilibrium phases of a ferromagnetic Ising spin Hamiltonian with nearest-neighbour interactions $-J{\sigma }_z^{(i)}{\sigma }_z^{(j)}$. At finite $N$, the all-up and all-down vectors belong to one Hilbert space, and the coherent stateis distinguishable from the corresponding mixture by a suitably chosen global observable. But if $A_R$ acts on only $R<N$ spins, its off-diagonal matrix element between these vectors vanishes: the untouched spins contribute $⟨\uparrow |\downarrow ⟩=0$. Hence its expectations already equal those of the equal-weight mixture, independent of $\theta$. Increasingly global observables can recover the phase at finite $N$, but the infinite product of spin flips is not a norm limit of finite-support observables and is absent from $\mathcal{A}$.

In the infinite-volume limit the up and down phases have disjoint state representations. The averaged magnetization $M_N=N^{-1}{\sum }_i{\sigma }_z^{(i)}$ commutes asymptotically with any fixed local observable: $\Vert [M_N,A_R]\Vert \le 2R\Vert A_R\Vert /N$. In the phase representations its limiting values are $+1$ and $-1$, distinguishing sectors. Finite spin-flip vectors form a dense set on which these averages tend to their respective scalar values, so boundedness extends the strong limits to the phase Hilbert spaces. An intertwiner between the representations would obey $(-I)T=T(+I)$ and must therefore vanish, explaining their disjointness. In the combined representation the observable algebra is block-diagonal between them, so no quasilocal measurement detects their relative phase. This yields an emergent superselection rule relative to the chosen observable algebra. It is not a derivation of physical wave-function collapse. Fractal dimension, a thermodynamic singularity and sector separation each exhibit new limiting structure while retaining an explicit microscopic construction.

Let $C$ be the middle-third Cantor set and choose a positive integer $d$ with $d\log 2/\log 3\ge k$; this is possible for every real $k$, including negative $k$. Set $X=C^d\subset {\mathbb{R}}^d$. It is closed and bounded since each factor is compact. If a subset of $X$ is connected, each coordinate projection is a connected subset of $C$ and hence a singleton; therefore the subset itself is a singleton. Thus $X$ is totally disconnected.

For completeness, the dimension of these Cartesian powers of the Cantor set can be checked directly. At stage $n$ there are $2^{dn}$ cubes of side $3^{-n}$, giving vanishing $s$-dimensional cover cost whenever $s>d\log 2/\log 3$. Conversely put independent equiprobable zero-or-two ternary digits on every coordinate. Each stage-$n$ cube has probability $2^{-dn}$. If $3^{-n}\le r<3^{-n+1}$, a Euclidean ball of radius $r$ meets at most a constant depending on $d$ such cubes. Hence its probability is at most $C_d{r}^{d\log 2/\log 3}$. For any sufficiently fine cover, summing this bound over covering sets shows that its cost at exponent $d\log 2/\log 3$ is bounded below by a positive constant, since total probability is one. These upper and lower Hausdorff measure bounds proveThe construction does not claim arbitrarily large Hausdorff dimension in one fixed ambient Euclidean dimension.

First we prove the Krein-Milman theorem in the required setting. If $K$ is empty the conclusion is immediate, so assume it is nonempty. A face of a convex set is a convex subset $F$ with the property that an interior point of a segment in $K$ belongs to $F$ only if both endpoints do. Consider all nonempty weakly compact faces of $K$, ordered by reverse inclusion. A chain has a nonempty intersection by compactness and the finite intersection property; that intersection is again a compact face. The Zorn lemma gives a minimal compact face $F$.

If $F$ contained distinct points, a bounded linear functional separating them would be nonconstant on $F$. Its maximizer set is a nonempty proper compact face of $F$, hence a face of $K$, contradicting minimality. Therefore $F$ is a singleton and its member is an extreme point of $K$. The same reasoning applies to each nonempty compact face of $K$, so each such face contains an extreme point of $K$.

Let $H$ be the norm-closed convex hull of the extreme points of $K$. A weakly compact subset of a Banach space is weakly closed, hence norm closed, so $H\subseteq K$. If $x\in K\setminus H$, the Hahn-Banach separation theorem gives $\ell \in E^{\prime }$ with $\ell (x)>{\sup }_H\ell$. The maximizer face of $\ell$ on $K$ contains an extreme point $e$ of $K$. Then $\ell (e)\ge \ell (x)>{\sup }_H\ell$, contradicting $e\in H$. We concludeNorm and weak closed convex hulls coincide by the same separation theorem. The printed phrase “closed convex cover” is understood in this standard closed-convex-hull sense.

Now work in the real space of continuous functions on a compact space $C(C)$ on the Cantor set. The extreme points of a real continuous-function unit ball are exactly the continuous sign functions:If $|f(x_0)|<1$, continuity supplies a neighbourhood where $|f|\le 1-\epsilon$. The Cantor cylinder sets form a clopen base, so choose a nonempty cylinder $D$ inside that neighbourhood. Its indicator function is continuous, and $f\pm \epsilon 1_D$ are distinct members of $B$ with midpoint $f$. Hence $f$ is not extreme. Conversely, if $f$ is pointwise sign-valued and $f=(g+h)/2$ with $g,h\in B$, equality at the endpoint of the scalar interval $[-1,1]$ forces $g(x)=h(x)=f(x)$ at every point.

To prove the closed-hull assertion without assuming weak compactness, take $f\in B$ and $\epsilon >0$. A sufficiently fine finite partition of $C$ into Cantor cylinders has oscillation of $f$ less than $\epsilon$ on each cell, by uniform continuity. Choose a value $a_j\in [-1,1]$ on each cell and let $h$ be the corresponding continuous step function. Then $\Vert f-h{\Vert }_{\infty }<\epsilon$. For each sign vector $s\in \{-1,1{\}}^m$, let $e_s$ take value $s_j$ on cell $j$, and assign the weightThese weights are nonnegative, sum to one, and satisfy ${\sum }_s{w}_s{s}_j=a_j$. Thus $h={\sum }_s{w}_s{e}_s$ is a convex combination of extreme points. This is clopen sign approximation in the real Cantor unit ball, provingNevertheless $B$ is not weakly compact. The suggested functions $f_n(x)=\min (3^{n}x,1)$ lie in $B$ and converge pointwise to the function equal to zero at $0$ and one at every other point of $C$. It is discontinuous at zero, since $2\cdot 3^{-j}\in C$ tends to zero. If $B$ were weakly compact, the sequence, viewed as a net, would have a weakly convergent subnet with limit in $C(C)$. Point evaluations are bounded linear functionals, so that subnet would have the same pointwise limit; its indices are cofinal in the original sequence. The limit would therefore be the discontinuous function just described, a contradiction. This discontinuous pointwise limit obstruction to weak compactness gives the required failure of weak compactness, despite the closed-hull equality.

The Rudolph measure rigidity theorem has an essential ergodicity hypothesis. If a Borel probability measure $\mu$ on the circle group is invariant under both $T_2$ and $T_3$, is ergodic for the semigroup generated jointly by these maps, and either map has positive Kolmogorov-Sinai entropy, thenEquivalently, a jointly ergodic common invariant measure other than Lebesgue measure has zero entropy for both maps. Joint ergodicity means that every set invariant modulo $\mu$ under both maps has measure zero or one. Positive entropy without this hypothesis is insufficient: $\frac{1}{2}m+\frac{1}{2}{\delta }_0$, with ${\delta }_0$ a Dirac measure, is a common invariant measure of positive entropy and is not $m$.

The Host equidistribution theorem states that if $p,q\ge 2$ are relatively prime integers and $\nu$ is invariant and ergodic under $T_p$, with $h_{\nu }(T_p)>0$, then for $\nu$-almost every $x$ the sequence $(T_q^{n}x)$ is an equidistributed sequence for Lebesgue measure. Explicitly, for every continuous $g$ on the circle,The non-ergodic form assumes $T_p$ invariance and positive entropy for almost every component in the ergodic decomposition $\nu =\int {\nu }_{\omega }d\lambda (\omega )$. Applying the ergodic theorem of Host on each such component gives the same almost-everywhere conclusion for $\nu$. More generally, its conclusion holds on the part supported on positive-entropy components. A positive value of $h_{\nu }(T_p)$ alone does not eliminate zero-entropy components.

To deduce the joint version of the Rudolph measure rigidity theorem, suppose $h_{\mu }(T_2)>0$; if only $T_3$ has positive entropy, interchange the roles. Write the $T_2$ ergodic decomposition as $\mu =\int \nu d\tau (\nu )$. Since $T_3$ commutes with $T_2$, its pushforward measure sends a $T_2$ ergodic component $\nu$ to a $T_2$ ergodic component $(T_3{)}_{\ast }\nu$. On each component, $T_3$ is a factor of a measure-preserving system with fibres of size at most three. We use the standard entropy preservation under a finite-to-one factor:The reason for this standard entropy fact is that, conditional on a complete factor point, every finite orbit name has at most three possibilities; its conditional entropy is bounded by $\log 3$, and division by the orbit length gives zero relative entropy.

The component at $T_{3}x$ is $(T_3{)}_{\ast }{\nu }_x$ almost everywhere; this follows from commutation and the componentwise ergodic averages. The component entropy function $r(x)=h_{{\nu }_x}(T_2)$ is therefore invariant under both $T_2$ and $T_3$. Joint ergodicity makes it constant almost everywhere, and affinity of entropy under ergodic decomposition identifies the constant as $h_{\mu }(T_2)>0$. Thus almost every $T_2$ component has positive entropy, exactly the condition required in the non-ergodic Host equidistribution theorem. It follows that $\mu$-almost every point equidistributes for $m$ under $T_3$.

For any continuous $g$, invariance under $T_3$ and the dominated convergence theorem now giveContinuous functions determine Borel probability measures on the circle, so $\mu =m$, proving the deduction.

For the normal-number example, let ${\omega }_1,{\omega }_2,\dots$ be independent fair binary digits and defineThis Cantor Bernoulli measure is supported on the middle-third Cantor set $C$. If $S$ is the Bernoulli shift, then $T_3\circ \pi =\pi \circ S$ on the circle. Consequently $\nu$ is $T_3$ invariant and ergodic: a $T_3$ invariant event pulls back to an $S$ invariant event, which has probability zero or one.

Take the ternary digit measurable partition $\eta =\{[0,1/3),[1/3,2/3),[2/3,1)\}$. Its block partition of length $n$ has, up to null endpoints, $2^n$ positive-measure atoms under $\nu$, each of measure $2^{-n}$, and all other atoms have measure zero. HenceApply the Host equidistribution theorem with $p=3$, $q=2$. For $\nu$-almost every $x\in C$, the sequence $(T_2^{n}x)$ equidistributes for Lebesgue measure, so $x$ is a normal number in base $2$ by normality and equidistribution under integer multiplication.

The ternary expansion of $\nu$-almost every such $x$ contains only $0$ and $2$, so the frequency of digit $1$ is zero rather than $1/3$. The ambiguous ternary endpoints form a countable $\nu$ null set and can be removed. Therefore $x$ is not a normal number in base $3$. We have proved the stronger almost-everywhere existence statement

Let $C_n$ be the union of the $2^n$ closed intervals remaining after $n$ steps of the Cantor set construction. Their total length is $(2/3{)}^n$. Any point outside $C_n$ has a ternary expansion with a $1$ among its first $n$ digits, so $A^c\subseteq C_n$. Endpoints of deleted intervals also belong to $A$ because they have an alternative ternary expansion containing a $1$, as in $2/3=0.1222\dots {}_3$.

A partition using all endpoints of $C_n$ has lower Darboux sum at leastSince $A$ is dense, every upper Darboux sum is one. Letting $n\to \infty$ gives equal upper and lower integrals, and therefore