Scalar proofs before the spectral construction

Original exposition and adaptation: GPT-6 Astra (OpenAI), Ultra, October 2026. CC0-1.0. The summable-subsequence argument and the six-step measure construction below develop the earlier programme proofs in Measure and Hilbert space tools for Haar integration, Section 3 (GPT-6.1 Sol, OpenAI, Ultra), and Haar measure on locally compact groups, Theorem 2.2 (originally Claude Opus 5.5, Anthropic; revised GPT-6.1 Sol, OpenAI, Ultra). Those earlier texts dedicate their original exposition under CC0. This version proves the compact-metric topology it needs and supplies the full scalar contracts used in the spectral lesson.

Read Scalar measure and integration: programme proofs, Sections 0–2 first. They prove the outer-measure construction, simple approximation, measurable limits, nonnegative integration, monotone convergence, Fatou's lemma and scalar dominated convergence, starting from elementary set theory and the complete ordered real field. On a noncomplete measure space, an almost-everywhere limit is taken with a measurable representative. All functions integrated below are measurable. No Hilbert-space representation theorem, spectral theorem, or general locally compact topology theorem is assumed in this note.

SS1. Scalar L2 on every measure space

Let $(X,\Sigma,\mu)$ be any measure space. A scalar function is identified with another when they agree outside a measurable null set. Let $L^2(\mu)$ be the classes of complex measurable functions with finite $\int|f|^2$. There is no finiteness, sigma-finiteness or completeness assumption on $\mu$.

For $f,g\in L^2$, the pointwise inequality $2|fg|\leq|f|^2+|g|^2$ proves that $f\bar g$ is integrable. Set $\langle f,g\rangle=\int f\bar g\,d\mu, \qquad \|f\|_2^2=\int |f|^2\,d\mu. \tag{SS.1}$ The integral properties in the preceding note prove sesquilinearity and conjugate symmetry. The pairing is linear in the first entry. If its diagonal is zero, then $\mu(\{|f|>1/n\})\leq n^2\int|f|^2=0$ for every positive integer $n$; the countable union shows $f=0$ almost everywhere. Thus it is positive definite on the quotient. For $g\ne0$, expand the nonnegative integral of $|f-zg|^2$ and choose $z=\langle f,g\rangle/\|g\|_2^2$. This gives $|\langle f,g\rangle|\leq\|f\|_2\|g\|_2. \tag{SS.2}$ For $g=0$ the inequality holds because $g$ vanishes almost everywhere. Expanding $\|f+g\|_2^2$ and using (SS.2) proves the triangle inequality. In particular all norm estimates used in the completeness proof have now been established directly.

Completeness. Given a Cauchy sequence $f_n$, choose a subsequence $f_{n_j}$ with $\|f_{n_{j+1}}-f_{n_j}\|_2\leq2^{-j}$, and choose measurable representatives. The partial sums of $v=\sum_{j\geq1}|f_{n_{j+1}}-f_{n_j}| \tag{SS.3}$ have norm at most $\sum_j2^{-j}=1$ by the triangle inequality. Monotone convergence applied to their squares gives $\int v^2\leq1$. Therefore $v$ is finite off a measurable null set: $\mu(\{v>m\})\leq m^{-2}$, and the set where $v=\infty$ lies in every such set. The subsequence consequently converges there to a scalar $f$. Define $f=0$ on the exceptional measurable set; the measurable-limit result makes $f$ measurable. Pointwise off that set, $|f-f_{n_j}|\leq\sum_{k\geq j}|f_{n_{k+1}}-f_{n_k}|.$ The finite-sum triangle inequality and then monotone convergence give $\|f-f_{n_j}\|_2\leq\sum_{k\geq j}2^{-k}. \tag{SS.4}$ This also proves $f\in L^2$, by adding the fixed first subsequence term. The subsequence converges in norm; the Cauchy property then gives norm convergence of the original sequence to the same limit. This proves that (SS.1) makes $L^2(\mu)$ a Hilbert space for every measure.

Simple approximation. A square-integrable representative can be chosen finite everywhere, by changing its infinite values on the measurable null set where they occur. For positive integers $n$, put $E_n=\{1/n<|f|\leq n\}$. These sets increase to $\{f\ne0\}$, and $\mu(E_n)\leq n^2\|f\|_2^2<\infty$. Dominated convergence gives $f1_{E_n}\to f$ in $L^2$. On $E_n$, rounding the real and imaginary parts to a sufficiently fine finite grid gives a simple function with pointwise error at most $\varepsilon/(1+\mu(E_n))^{1/2}$. Its squared-integral error is at most $\varepsilon^2$. Set it to zero off $E_n$. Thus simple functions supported on finite-measure sets are dense. The same union $\bigcup_nE_n$ shows that every $L^2$ function is supported on a sigma-finite measurable set, even when the whole measure space is not sigma-finite.

SS2. The topology needed on a compact metric space

Let $K$ be a compact metric space. A closed subset is compact: add its open complement to an open cover and use compactness of $K$. A compact subset is closed: for a point outside it, finitely many disjoint metric neighborhoods of that point and points of the compact set give a separating open neighborhood. Continuous images of compact sets are compact, since inverse images carry open covers back to the domain. A continuous real function on a nonempty compact set is bounded, as finitely many inverse images of bounded intervals cover it. It attains its supremum: the closed sets where its value is at least the supremum minus $1/n$ have the finite-intersection property, and an empty intersection would contradict compactness by their open complements. Infima follow by negation.

For nonempty $F\subseteq K$, the function $d(x,F)=\inf_{y\in F}d(x,y)$ is 1-Lipschitz, by applying the triangle inequality and taking infima in both directions. If $F$ is closed, this distance vanishes exactly on $F$.

If $F\subset U\subseteq K$, with $F$ compact and $U$ open, there is a continuous $g:K\to[0,1]$ equal to one on $F$ with $\operatorname{supp}g\subset U$. When $F=\varnothing$ use zero; when $U=K$ use one. Otherwise the function $d(x,K\setminus U)$ has a strictly positive minimum $a$ on $F$. The formula $g(x)=\max(0,1-2d(x,F)/a) \tag{SS.5}$ has the required properties: its support lies in $\{d(x,F)\leq a/2\}$, whereas every point of $K\setminus U$ has distance at least $a$ from $F$. The same argument gives an open $V$ with $F\subset V\subset\overline V\subset U$, by taking $V=\{d(x,F)<a/3\}$. The empty and whole-space cases are handled as above.

We also need a finite partition. Suppose the compact set $F$ lies in $U_1\cup\cdots\cup U_n$. For each $x\in F$, choose one $U_i$ containing it and a metric ball about $x$ whose closed ball of twice the radius lies in $U_i$. Finitely many of the smaller balls cover $F$. For each $i$, let $F_i$ be the union of their closed smaller balls assigned to $i$, or the empty set if none were assigned. These are compact subsets of $U_i$, and cover $F$. Choose $g_i$ as in (SS.5), equal to one on $F_i$ and supported in $U_i$. Put $h_1=g_1,\qquad h_i=g_i\prod_{j<i}(1-g_j)\quad(i>1). \tag{SS.6}$ Each $h_i$ is continuous, lies between zero and one, and is supported in $U_i$. Induction gives $\sum_ih_i=1-\prod_i(1-g_i)$. The sum is at most one everywhere and is one on $F$. This proves the complete partition assertion used below.

SS3. A positive functional is a finite regular measure

Theorem. Every positive real-linear functional $I:C(K,\mathbb R)\to\mathbb R$ on a compact metric space is represented by a unique finite regular Borel measure $\mu$: $I(f)=\int_K f\,d\mu, \qquad\mu(K)=I(1). \tag{SS.7}$ Regularity here includes outer approximation by open sets and inner approximation by compact sets for every Borel set. For the empty space the measure and functional are zero, so assume $K\ne\varnothing$. Positivity implies monotonicity and $|I(f)|\leq I(1)\|f\|_\infty$, by bounding $f$ between its two constant norm bounds.

Write $f\prec U$ when $f$ is continuous, $0\leq f\leq1$, and $\operatorname{supp}f\subset U$, with $U$ open in $K$. All supports are compact. We give the measure construction and representation in full.

Step 1: uniqueness from open and compact sets. If a finite regular measure represents $I$, then $\mu(U)=\sup_{f\prec U}I(f),\qquad \mu(F)=\inf_{1_F\leq f\in C(K,\mathbb R)}I(f) \quad(F\text{ compact}). \tag{SS.8}$ For the first identity, every displayed integral is at most $\mu(U)$. For each compact $F\subset U$, SS2 supplies such an $f$ equal to one on $F$, whose integral is at least $\mu(F)$; inner regularity proves equality. For the second, $1_F\leq f$ implies $f\geq0$ everywhere and gives the lower bound. For any open $U\supset F$, the same cutoff has integral at most $\mu(U)$; outer regularity gives equality. The first formula determines the measure on open sets, and outer regularity determines it on every Borel set. This proves uniqueness once existence is established.

Step 2: an outer measure. Define $m(U)=\sup_{f\prec U}I(f),\qquad m^*(A)=\inf_{U\supset A,\ U\text{ open}}m(U) \quad(A\subset K). \tag{SS.9}$ These numbers lie in $[0,I(1)]$. Monotonicity gives $m^*(U)=m(U)$ for open $U$, and $m^*(\varnothing)=0$. If $U=\bigcup_jU_j$ and $f\prec U$, finitely many $U_j$ cover its compact support. The partition in SS2 gives continuous $h_j\prec U_j$ with $f=\sum_jfh_j$. Each $fh_j\prec U_j$, so $I(f)\leq\sum_jm(U_j)$; taking the supremum gives $m(U)\leq\sum_jm(U_j)$. For arbitrary sets $A_j$, choose open $U_j\supset A_j$ with $m(U_j)<m^*(A_j)+\varepsilon2^{-j}$. The open-set inequality gives $m^*(\bigcup_jA_j)\leq\sum_jm^*(A_j)+\varepsilon$. Let $\varepsilon\downarrow0$. Thus $m^*$ is an outer measure.

Step 3: Borel measurability. Fix open $U$ and first let $V$ be open. Choose $f\prec V\cap U$ with $I(f)>m(V\cap U)-\varepsilon$, and then $g\prec V\setminus\operatorname{supp}f$ with $I(g)>m(V\setminus\operatorname{supp}f)-\varepsilon$. Their supports are disjoint, so $f+g\prec V$. Because $V\setminus U\subset V\setminus\operatorname{supp}f$, $m(V)>m^*(V\cap U)+m^*(V\setminus U)-2\varepsilon. \tag{SS.10}$ For an arbitrary $A$, choose open $V\supset A$ with $m(V)<m^*(A)+\varepsilon$, and use monotonicity in (SS.10). Letting $\varepsilon\downarrow0$ proves the greater-than-or-equal half of the Carathéodory condition for $U$; subadditivity gives the other half. The full Carathéodory theorem proved in the preceding programme note, Section 1, now makes the Borel sets measurable and the restriction $\mu=m^*|_{\mathcal B(K)}$ a measure. It is finite and outer regular by construction.

Step 4: compact sets. If $F$ is compact and $1_F\leq f\in C(K,\mathbb R)$, then for $0<c<1$ the open $U_c=\{f>c\}$ contains $F$. Every $g\prec U_c$ satisfies $g\leq f/c$, so $\mu(F)\leq\mu(U_c)\leq I(f)/c$. Let $c\uparrow1$. Conversely, outer regularity provides open $U\supset F$ with $\mu(U)<\mu(F)+\varepsilon$. A cutoff equal to one on $F$, supported in $U$, satisfies $I(f)\leq m(U)=\mu(U)$. This proves the second formula in (SS.8) for the constructed measure.

Step 5: inner regularity. If $f\prec U$, then $I(f)\leq\mu(V)$ for every open $V\supset\operatorname{supp}f$. Outer regularity gives $I(f)\leq\mu(\operatorname{supp}f)$. Taking the supremum over $f\prec U$ proves $\mu(U)\leq\sup_{F\subset U,\ F\text{ compact}}\mu(F)$; monotonicity gives the reverse inequality. For a Borel $E\subset K$, choose open $O\supset K\setminus E$ with $\mu(O)<\mu(K\setminus E)+\varepsilon$. Then the compact set $F=K\setminus O$ lies in $E$, and finiteness gives $\mu(E\setminus F)=\mu(O)-\mu(K\setminus E)<\varepsilon$. Thus every Borel set has compact inner approximation. Along with outer regularity, this supplies compact $F\subset E\subset U$ open with $\mu(U\setminus F)<\varepsilon$, after choosing both errors less than $\varepsilon/2$.

Step 6: representation. It suffices by real linearity, scaling and positive/negative parts to treat continuous $0\leq f\leq1$. For a positive integer $N$, set $F_0=\operatorname{supp}f,\quad F_j=\{f\geq j/N\}\ (1\leq j\leq N), \qquad f_j=\min(\max(f-(j-1)/N,0),1/N). \tag{SS.11}$ Then $f=\sum_{j=1}^Nf_j$, $\operatorname{supp}f_j\subset F_{j-1}$, and $1_{F_j}/N\leq f_j\leq1_{F_{j-1}}/N$. Integration bounds $\int f_j$ by the corresponding two masses. Step 4 gives $\mu(F_j)\leq I(Nf_j)$. For every open $V\supset F_{j-1}$, one has $Nf_j\prec V$, so outer regularity gives $I(Nf_j)\leq\mu(F_{j-1})$. Therefore both $I(f)$ and $\int f\,d\mu$ lie in the interval $\left[\frac1N\sum_{j=1}^N\mu(F_j),\ \frac1N\sum_{j=0}^{N-1}\mu(F_j)\right], \tag{SS.12}$ whose length is $(\mu(F_0)-\mu(F_N))/N\leq\mu(K)/N$. Let $N\to\infty$. This proves (SS.7) for every real continuous function. Taking the constant one gives the mass formula. The uniqueness in Step 1 completes the proof.

For a complex continuous function $f=u+iv$, the complexification $I_{\mathbb C}(f)=I(u)+iI(v)$ consequently equals its complex integral. In particular, a vector functional of a unital *-representation gives exactly the complex integral identity used in SK02 after its positivity has been checked there.

Four layers bound the same continuous integral from below and above

Figure. For the exact example $K=[0,1]$, $f(x)=x$, ordinary length measure and $N=4$, the lower and upper step functions in (SS.12) have integrals $3/8$ and $5/8$. Both the functional value and the integral are trapped in that interval; its width is $1/4$. Step 6 proves that the width tends to zero for every continuous function and every finite measure constructed here. The plotted values at individual subdivision endpoints do not affect this length-measure example; the proof retains their exact indicator inequalities even when a measure has atoms.

SS4. Completion of a Borel measure does not change scalar L2

Let $\mu$ be a Borel measure on $K$, and $\bar\mu$ its completion. By the definition of completion, a completion-measurable set differs from a Borel set by a subset of a Borel null set. If $f$ is a complex completion-measurable function, choose simple completion-measurable $s_n\to f$ pointwise by truncating and rounding a finite complex grid. Replace each of their finitely many level sets by a Borel representative and keep the same finite linear combination of indicators. The resulting Borel function $t_n$ agrees with $s_n$ off a Borel null set $N_n$; possible overlaps of the representatives do not affect that assertion.

The union $N=\bigcup_nN_n$ is Borel and null. Off $N$, the sequence $t_n$ converges to $f$. Define $g$ to be that limit off $N$ and zero on $N$. Its measurability follows by applying the measurable-limit result to $t_n1_{K\setminus N}$, a Borel sequence converging everywhere. Thus $g$ is Borel and equals $f$ almost everywhere. Conversely every Borel function is completion-measurable, and its integrals are unchanged, first for simple functions and then by monotone convergence. Hence the quotient $L^2$ spaces, their pairings and their norms are canonically identical. No arbitrary modification on a nonmeasurable subset of a null set is called Borel.

SS5. Exact inputs supplied

The preceding programme note, Sections 0–2, supplies nonnegative integration, simple approximation, Carathéodory, monotone convergence, Fatou and scalar dominated convergence on arbitrary measure spaces. SS1 supplies the linear-first complex Hilbert structure and completeness of scalar $L^2$ for every measure. SS2–SS3 give positive real functional representation on every compact metric space, with unique finite regular Borel measure and compact inner approximation of every Borel set. SS4 supplies the completed-measure representative used in the cyclic representation. These are the full scalar hypotheses of SK01–SK05; none assumes separability of the operator's Hilbert space.

This note is an adaptation with newly written proofs, rather than a transfer of the earlier editions' review verdicts. It does not certify the broader locally compact group lesson or its other topology prerequisites. The spectral lesson still proves its own continuous-density, arbitrary-direct-sum and fixed-vector measure bridges. The proofs in this note and the preceding measure note, together with those bridges, supply the written scalar route.