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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.

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.

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.

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.