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A binary tree of finite strings is a set of finite binary strings closed under taking prefixes. The empty tree is allowed; a nonempty such tree contains the empty string. Finite nodes may have no infinite extensions. Such presentations describe closed subsets of Cantor space through their infinite paths.

Given a continuous surjection $h:\Omega \to K$ from Cantor space to a nonempty compact metric space, the pullback $f\mapsto f\circ h$ is a unital isometric embedding of $C(K)$ into $C(\Omega )$. Transport a normalized positive linear functional to its image, take a positive extension from a unital subspace of C(K), and represent that extension on Cantor space. The pushforward measure $h_{\ast }P$ then represents the original functional.

A continuous function on Cantor space is uniformly continuous. Replacing it by one constant on each length-$n$ prefix cylinder set changes it by at most its oscillation on sets of diameter $2^{-n}$. This tends uniformly to zero. Each approximant is a finite linear combination of continuous indicator functions, proving density in the supremum norm.

A normalized positive linear functional on $C(\Omega )$ for Cantor space gives a finitely additive set function on the algebra of clopen sets. If a countable disjoint union of such sets is itself clopen, compactness gives a finite subcover; all other terms are empty. Finite additivity therefore gives the required countable additivity, making $\ell$ a premeasure. The Caratheodory extension theorem produces a unique Borel probability measure.

Every open set in Cantor space is a countable disjoint union of prefix cylinder sets. Select all finite words whose cylinder lies in the open set but whose immediate predecessor's cylinder does not. These words are prefix-free, so their cylinders are disjoint. Every point of the open set has a shortest qualifying prefix. The whole space is represented by the empty prefix.

An infinite path is a binary function all of whose finite prefixes belong to the tree. Its path space $[T]$ is closed in Cantor space, since failure is witnessed by one finite prefix outside $T$.

Take $\mathbb{N}=\{0,1,2,\dots \}$ and fix a standard effective enumeration $({\phi }_e)$ of the unary partial computable functions by program texts. WriteThe diagonal halting set $K$ is not a computable set: a supposed decision procedure would give a program that halts on input $e$ exactly when $e\notin K$, contradicting its own index. Each $K_s$ is finite and uniformly decidable by bounded simulation of a computation, and $K_s\subseteq K_{s+1}$ with union $K$.

For $a<b<c$, define the computable colouringThis is a two-part decidable set partition on three-element subsets: sort the three inputs and perform the two finite simulations. Here $[0,a)=\{0,\dots ,a-1\}$. The halting-stage colouring of triples cannot have an infinite homogeneous set for a colouring of colour $1$. Indeed, fix its least element $a$. The finitely many bits of $K_s\cap [0,a)$ eventually stabilize. Choose two later elements $b<c$ of the putative homogeneous set for a colouring beyond that stabilization stage; the corresponding triple has colour $0$.

Now suppose $H$ is an infinite homogeneous set for a colouring of colour $0$. Its membership oracle computes $K$: to decide whether $n\in K$, search for $a,b\in H$ with $n<a<b$, and return whether $n\in K_b$. For every $c\in H$ above $b$, homogeneity gives $K_c\cap [0,a)=K_b\cap [0,a)$. Since $H$ is unbounded, taking such $c$ arbitrarily large proves that this common finite set is $K\cap [0,a)$. ThusIf $H$ were a computable set, the displayed Turing reduction would decide $K$, a contradiction. The displayed decidable two-colouring has no infinite decidable homogeneous set.

For the second construction, let $2^{<\omega }$ be the finite binary strings, with coordinates starting at $0$, and define the computable binary treeMembership is decided by finitely many tests using bounded simulation of a computation. It is a binary tree of finite strings: if a string satisfies the tests, each prefix satisfies its shorter and fewer tests. The infinite paths through a binary tree are exactlyThis set is nonempty: at a coordinate whose diagonal computation halts with a binary value choose the other value, and choose either bit elsewhere. This is an existence argument, not a proposed computable decision procedure for convergence. No member is a total computable function. If $f={\phi }_d$ were a total binary-valued function, the condition at $d$ would say $f(d)\ne {\phi }_d(d)=f(d)$. Thus each member is a binary diagonally noncomputable function.

Moreover, $[T]$ has no isolated point. There are arbitrarily large indices of programs that never halt: for example, distinct program texts with successively more unused instructions followed by an infinite loop provide infinitely many such indices in the usual syntactic program coding. At every one of these coordinates the bit is unrestricted. Given any $f\in [T]$ and any prefix length $n$, change its bit at one such index $e\ge n$ and leave all other bits unchanged. The resulting different path still belongs to $[T]$ and has the same length-$n$ prefix. Since a path space is closed in Cantor space, $T$ is decidable, $[T]$ is nonempty and perfect, and no infinite path is computable.

There is a definition issue with the adjective “perfect”. The preceding conclusion uses perfect path space, allowing finite dead ends in the decidable presentation. If “perfect tree” instead requires every node to have two incompatible extensions in the tree, the requested combination is impossible. Such a nonempty tree has no terminal nodes. Starting at the empty string, test the two immediate successors and choose the first that belongs to the tree; one always exists. This constructs a total computable function whose successive prefixes remain in the tree. Thus computable pruned binary trees have computable paths. Our $T$ necessarily has dead ends; its subtree consisting only of prefixes of infinite paths is perfect in the pruned sense but is not decidable. The construction therefore supplies the intended perfect closed path space and also resolves the stronger, inconsistent interpretation.

Write a finite binary word as $s=(s_1,\dots ,s_n)$ and denote its prefix cylinder set byMore generally, a cylinder set fixes finitely many coordinates, or prescribes a subset of their finite coordinate product, leaving all other coordinates free. Each coordinate factor is discrete, so finite-coordinate inverse images are both open and closed. Thus cylinders are clopen. The basic open sets of the product topology restrict only finitely many coordinates, so cylinders form a base. Any finite-coordinate cylinder is a finite union of prefix cylinders of one sufficiently long common length; the prefix cylinders therefore also form a base.

The Bernoulli space here is Cantor space, with compatible metricIt is a compact metric space: from any sequence, choose successively subsequences constant in the first, second and subsequent coordinates, and take the diagonal subsequence. Agreement on the first $n$ coordinates bounds the distance by $2^{-n}$, proving convergence. Compactness also follows from Tychonoff theorem.

Every cylinder indicator function is continuous because the cylinder is clopen. Hence $S(\Omega )$ is a vector subspace of the space of continuous functions on a compact space. To prove density, fix $f\in C_{\mathbb{R}}(\Omega )$ and $\epsilon >0$. By uniform continuity, choose $n$ so that $d(x,y)\le 2^{-n}$ implies $|f(x)-f(y)|<\epsilon$. Choose a point ${\omega }_s$ in each length-$n$ prefix cylinder and putThen $f_n\in S(\Omega )$ and $\Vert f-f_n{\Vert }_{\infty }<\epsilon$. This proves cylinder-function density in Cantor space:

Positivity of $\varphi$ implies monotonicity. The pointwise bounds $-\Vert f{\Vert }_{\infty }1\le f\le \Vert f{\Vert }_{\infty }1$ therefore giveEvaluation on $1$ attains equality, so $\varphi$ is continuous and has norm one. This is the norm of a positive functional on C(K).

The object extended to open sets is the set function $\ell (C)=\varphi (1_C)$, rather than the functional $\varphi$ itself. Positivity gives $\ell (C)\ge 0$, and linearity gives finite additivity on disjoint cylinders. To construct its extension directly, use the disjoint cylinder decomposition of open subsets of Cantor space. For an open $U$, take the shortest prefixes $s$ for which $[s]\subset U$. They are prefix-free, their cylinders are disjoint, and their union is $U$. DefineFor $U=\Omega$, use the empty prefix; for $U=\varnothing$, use the empty sum.

This value is independent of the chosen disjoint prefix-cylinder decomposition. Indeed, compare two such decompositions $(C_i)$ and $(D_j)$. For each fixed $C_i$, its intersections with the $D_j$ give a disjoint open cover of the compact cylinder $C_i$. Compactness reduces this to finitely many nonempty intersections. Each intersection is a prefix cylinder or is empty, so finite additivity gives $\ell (C_i)={\sum }_j\ell (C_i\cap D_j)$. Sum over $i$ and rearrange the nonnegative double sum. Doing the same with each $D_j$ proves equality of the two totals.

For disjoint open sets $U_k$, combine their disjoint cylinder decompositions to obtain one for ${\bigcup }_k{U}_k$. Rearrangement of nonnegative sums proves countable additivity on this family of open sets. The extension is unique because any countably additive extension must have the prescribed sum on every disjoint cylinder decomposition. Open sets are not themselves a sigma-algebra; countable additivity here concerns disjoint open families and their open union.

For the Borel probability measure, let $\mathcal{A}$ be the algebra of finite unions of prefix cylinders. It is also the algebra of clopen sets, because every clopen set is compact and has a finite cylinder cover. Define $\ell (A)=\varphi (1_A)$ on $\mathcal{A}$. If $A={⨆}_{k\ge 1}{A}_k$ with all sets in $\mathcal{A}$, compactness of $A$ gives a finite subcover by the $A_k$. Disjointness makes all remaining members empty. Thus finite additivity already establishes the premeasure condition.

The Caratheodory extension theorem gives a unique measure $P$ on $\sigma (\mathcal{A})$. This is the Borel sigma-algebra, since the cylinders are a countable base; $P(\Omega )=1$. Its values on open sets agree with the extension above. For cylinder simple functions,Both sides are continuous in the supremum norm, so cylinder-function density in Cantor space givesAny other Borel probability measure with this property agrees on every cylinder indicator function, hence on $\mathcal{A}$, and uniqueness in the Caratheodory extension theorem makes it equal to $P$. This proves the Cantor-space representation of positive functionals without assuming the general representation theorem.

Now let $h:\Omega \to X$ be the supplied continuous surjection. The pullbackis a unital isometric embedding: surjectivity gives $\Vert f\circ h{\Vert }_{\infty }=\Vert f{\Vert }_{\infty }$. On the vector subspace $J(C_{\mathbb{R}}(X))$, define $L(Jf)=\varphi (f)$. This is well-defined, has norm one and satisfies $L(1)=1$.

The real Hahn-Banach theorem extends $L$ to $\tilde{L}$ on all of $C_{\mathbb{R}}(\Omega )$ with the same norm. Norm preservation alone does not automatically mean positivity, so verify it. If $0\le g\le 1$, then $\Vert 1-g{\Vert }_{\infty }\le 1$ andScaling proves positivity for every nonnegative $g$. This is the unital contraction positivity criterion, giving a positive extension from a unital subspace of C(K).

Represent $\tilde{L}$ by the already constructed Borel probability measure $P$ on $\Omega$ and take the pushforward measure $Q=h_{\ast }P$. Continuity of $h$ makes its Borel inverse images measurable, and $Q(X)=1$. The pushforward integral identity givesThis is compact-metric representation by Cantor-space pullback. The hypothesis $\varphi (1)=1$ already excludes an empty $X$. The measure is on the Borel sets; no uniqueness claim on an unspecified larger collection of subsets is needed.

A nonempty path space $[T]$ is perfect if it has no isolated points: for every path and every finite prefix of it there is a different path sharing that prefix. This property of the closed subset of Cantor space does not imply that its particular finite-string presentation has no dead ends. Requiring every finite node to extend to incompatible nodes is the stronger pruned-tree convention for a perfect tree.