Solution
= Solution
Write a finite binary word as $s=(s_1,\ldots,s_n)$ and denote its prefix <cylinder set> by
$$
[s]=\{\omega:\omega_1=s_1,\ldots,\omega_n=s_n\}.
$$
More 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 <metric>
$$
d(\omega,\omega')=\sum_{j\ge1}2^{-j}|\omega_j-\omega_j'|.
$$
It 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 $\varepsilon>0$. By <uniform continuity>, choose $n$ so that $d(x,y)\le2^{-n}$ implies $|f(x)-f(y)|<\varepsilon$. Choose a point $\omega_s$ in each length-$n$ prefix cylinder and put
$$
f_n=\sum_{s\in\{0,1\}^n}f(\omega_s)1_{[s]}.
$$
Then $f_n\in S(\Omega)$ and $\|f-f_n\|_\infty<\varepsilon$. This proves <cylinder-function density in Cantor space>:
$$
\boxed{\overline{S(\Omega)}^{\|\cdot\|_\infty}=C_{\mathbb R}(\Omega).}
$$
Positivity of $\phi$ implies monotonicity. The pointwise bounds $-\|f\|_\infty1\le f\le\|f\|_\infty1$ therefore give
$$
|\phi(f)|\le\|f\|_\infty\phi(1)=\|f\|_\infty.
$$
Evaluation on $1$ attains equality, so \b[$\phi$ 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 $\ell(C)=\phi(1_C)$, rather than the functional $\phi$ itself. Positivity gives $\ell(C)\ge0$, 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 $U$. Define
$$
\boxed{\ell(U)=\sum_{[s]\text{ in this decomposition}}\ell([s]).}
$$
For $U=\Omega$, 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 $\bigcup_kU_k$. 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 $\ell(A)=\phi(1_A)$ on $\mathcal A$. If $A=\bigsqcup_{k\ge1}A_k$ 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>,
$$
\phi(f_n)=\int_\Omega f_n\,dP.
$$
Both sides are continuous in the <supremum norm>, so <cylinder-function density in Cantor space> gives
$$
\boxed{\phi(f)=\int_\Omega f\,dP\qquad(f\in C_{\mathbb R}(\Omega)).}
$$
Any 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 pullback
$$
J:C_{\mathbb R}(X)\longrightarrow C_{\mathbb R}(\Omega),\qquad Jf=f\circ h
$$
is a unital <isometric embedding>: surjectivity gives $\|f\circ h\|_\infty=\|f\|_\infty$. On the <vector subspace> $J(C_{\mathbb R}(X))$, define $L(Jf)=\phi(f)$. This is well-defined, has norm one and satisfies $L(1)=1$.
The real <Hahn-Banach theorem> extends $L$ to $\widetilde L$ on all of $C_{\mathbb R}(\Omega)$ with the same norm. Norm preservation alone does not automatically mean positivity, so verify it. If $0\le g\le1$, then $\|1-g\|_\infty\le1$ and
$$
\widetilde L(g)=1-\widetilde L(1-g)\ge1-|\widetilde L(1-g)|\ge0.
$$
Scaling 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 $\widetilde L$ by the already constructed <Borel probability measure> $P$ on $\Omega$ and take the <pushforward measure> $Q=h_*P$. Continuity of $h$ makes its Borel inverse images measurable, and $Q(X)=1$. The pushforward integral identity gives
$$
\boxed{\phi(f)=\widetilde L(f\circ h)
=\int_\Omega f\circ h\,dP=\int_Xf\,dQ.}
$$
This is <compact-metric representation by Cantor-space pullback>. The hypothesis $\phi(1)=1$ already excludes an empty $X$. The <measure> is on the Borel sets; no uniqueness claim on an unspecified larger collection of subsets is needed.