Constant CAR states and the periodic core

Written by GPT-6.1 Sol (OpenAI), Ultra, October 2026. Author self-checked relative to the stated prerequisites; not independently reviewed. New original text: public domain (CC0).

An infinite sequence of identical two-level density matrices gives a type III factor when the two eigenvalues differ. To find its parameter, we must compute a modular invariant. A period of one modular action, or even its complete group of inner times, does not by itself justify the subtype.

We use Infinite tensor products and their reference states, especially its explicit faithful product representation, local slice maps and closed modular graph, and Fermions, Fock space and quasi-free factors, including its direct constant-state type III proof in Section 20. Sections 4–5 below construct the represented period-circle crossed product and continuous core. Scalar Lebesgue integration and Hilbert completion underlie the Fourier and measurable-phase arguments.

The modular foundations can be followed through BG1–6, which prove normalization, fullification and the complete closed modular graphs for an arbitrary left Hilbert algebra. FC1–5 prove fixed-corner GNS and modular restriction for every faithful normal semifinite weight, even when the corner has infinite weight. These arguments supply the pure Hilbert-algebra and fixed-corner steps in the MG comparison. The finite-ideal graph closedness and every-weight comparison retain their stated prerequisite proofs. Type III corner transport is a separate prerequisite of the general reverse inclusion in MG-4; the product-spectrum upper bound below does not use it.

For the subtype inference, Section 6 uses the full finite-ideal spectral comparison and the every-weight fixed-corner comparison in MG-0–3. PP00–PP04 provide the concrete filter and factor-contact arguments showing that every integer modular degree survives in every nonzero fixed corner. We also compute the entire spectrum of the product modular operator from its coefficient domain. These two calculations determine the intersection over all faithful normal semifinite weights, including weights of infinite total mass. The center calculation then gives, for this factor with least positive center period $L$,

Here (1) is a consequence of the all-weight calculation, not a premise inferred from periodicity. The upper bound on the intersection comes from the explicit product modular operator; its lower bound comes from comparison with every weight. The general type III corner-transport step of MG-4 is therefore unnecessary for this particular upper bound. Its full stated proof remains available for the general comparison problem.

The periodic subtype calculation fixes a separable one-particle Hilbert space. The subtype calculation below assumes this space is infinite-dimensional. The earlier CAR lesson separately retains finite-dimensional and arbitrary directed finite-mode approximation statements.

1. The constant faithful product state

Let $0<p<1$, $p\ne\tfrac12$, and set

On a separable infinite-dimensional one-particle space, the CAR parity strings identify the state with

This is an identification of the completed GNS representations, not just of their finite moments. Write $A_n=\bigotimes_{j=1}^nM_2\subset M$ and let $\Omega$ be the product GNS vector.

The state $\phi$ is faithful and normal. For clarity, in one coordinate represent $M_2$ by left multiplication on its Hilbert–Schmidt space with reference vector $\rho^{1/2}$. Since $\rho$ is invertible, this vector is cyclic both for left and for right multiplication. Finite left products applied to $\Omega$ span a dense subspace of the pointed product Hilbert space. Finite right products commute with the represented algebra and have the same density property. Thus $\Omega$ is cyclic for $M$ and for $M'$, hence separating for $M$. Its vector state is faithful.

Factorality follows from the concrete expectations constructed in Section 2. If $z\in Z(M)$, bimodularity puts $E_n(z)$ in $Z(A_n)=\mathbb C1$; state preservation makes it $\phi(z)1$. Lemma 2.1 gives $z=\phi(z)1$. The same lemma proves that the increasing finite matrix algebras have strong closure $M$, so $M$ is AFD.

For the modular calculation, choose in one Hilbert–Schmidt leg the orthonormal basis

Finite tensor words in these vectors, with the reference vector in all remaining legs, form an orthonormal basis $(\xi_\kappa)$ of the pointed completion. Put $m_\kappa=\#e_{12}-\#e_{21}$ and $d_\kappa=((1-p)/p)^{m_\kappa}$. The coefficient operator

is positive and selfadjoint: its resolvent at a nonreal number is the bounded coefficient inverse, whose image satisfies the displayed domain condition. Let $J$ conjugate coefficients and interchange $e_{12}$ with $e_{21}$ in each word. On a finite matrix vector, $J\Delta^{1/2}(a\Omega)=a^*\Omega$. The entire domains are

Testing the adjoint identity on each basis vector forces the coefficients of the antilinear adjoint, and their square summability gives exactly the second domain. Thus $(J\Delta^{1/2})^*=J\Delta^{-1/2}$. Membership in the product domain additionally requires $\sum d_\kappa^2|c_\kappa|^2<\infty$, and gives $(J\Delta^{1/2})^*(J\Delta^{1/2})=\Delta$ on its full displayed domain. Truncating coefficient series converges simultaneously in vector norm and in each corresponding weighted norm, so finite matrix vectors are graph cores. The full graph argument in Section 11 of the tensor-product lesson, using $E_n(x)\Omega\to x\Omega$ and $E_n(x^*)\Omega\to x^*\Omega$, identifies $J\Delta^{1/2}$ with the closure of $x\Omega\mapsto x^*\Omega$ on all of $M\Omega$. Indeed these approximating finite graph points converge to every initial graph point, while the finite-support core already closes to $J\Delta^{1/2}$; the two closed graphs therefore agree. Thus this $\Delta$ is $\Delta_\phi$, rather than an operator specified only on an algebraic core. Its imaginary powers fix $\Omega$, are strongly continuous by bounded coefficient convergence, and implement, on every local algebra,

Here $x\in A_n$ and $\rho_n=\rho^{\otimes n}$. Put

In particular $0<\lambda<1$ and $\omega=|\log\lambda|$.

If $b_j=e_{12}$ in coordinate $j$, then

Consequently $\sigma_P=\operatorname{id}$. This is the least positive period of the action: (5) forces $t\log r\in2\pi\mathbb Z$ at any period. For each integer $k$, there is a nonzero local operator $x_k$ with

Choose $b_j$ or $b_j^*$ with positive frequency $\omega$, and multiply $k$ copies in distinct coordinates when $k>0$; use adjoints for negative $k$, and $x_0=1$. Products in distinct tensor coordinates are nonzero.

For comparison with the scalar overlap appearing in the tensor-product lesson, at any $t$ for which the phases of $(1-p)^{it}$ and $p^{it}$ differ one has

Indeed, for $a=1-p$, $b=p$, $|z|=|w|=1$, $|az+bw|^2=1-ab|z-w|^2$. This proves (7), but a divergence criterion is not the type III argument used here.

To prove type III, let $X_j$ be a matrix of bounded norm in a tail leg. Right multiplication by $\rho^{-1/2}X_j\rho^{1/2}$ gives a commuting operator $R_j$ with $R_j\Omega=X_j\Omega$ and $\|R_j\|\le C(p)\|X_j\|$, where $C(p)$ is independent of $j$. Given $u\in M$, approximate $u\Omega$ by a local vector $a\Omega$. For $j$ outside the support of $a$,

The commutator commutes with the right local algebra and has a uniform norm bound. Density of its right orbit extends this limit to every vector. Repeating it with adjoints proves $[X_j,u]\to0$ strongly*. Apply this to the tensor matrix unit $b_j$ of (5), with its CAR parity string already included in the identification (2).

The projections $b_j^*b_j$ and $b_jb_j^*$ converge weakly to $p1$ and $(1-p)1$, respectively. Between two finite-support vectors, this is an exact equality once $j$ is beyond their supports; their uniform bounds extend it to arbitrary vectors.

Suppose that $e\in M$ is a nonzero finite projection. Its corner is a finite factor: the contact argument in Sections 8.5 and 11.3 proves factoriality of every nonzero corner of a factor. On the separable Hilbert space $eH$, Sections 18–19 of the CAR lesson construct a faithful normal tracial state $\tau$ on this finite factor, with $\tau(e)=1$. This is the supplied finite-factor trace proof, including normality through summable vector coefficients; it is not an infinite-product classification criterion. Set $B_j=eb_je$. Strong* commutator convergence gives

strongly: for example $(1-e)b_je=(1-e)[b_j,e]e\to0$, and multiplying by the uniformly bounded remaining factors preserves the limit. Hence $B_j^*B_j\to pe$ and $B_jB_j^*\to(1-p)e$ weakly. Normality passes these bounded limits through $\tau$: first pass each vector coefficient, then control the tail of its summable coefficient representation uniformly. Traciality gives $p=1-p$, contradicting the hypothesis. Thus $M$ has no nonzero finite projection and is type III. The complete GNS identification in the CAR lesson transports this conclusion to the CAR factor.

2. Slice expectations and the occupation diagonal

Regroup the represented product as $M=A_n\bar\otimes M_{>n}$, using the head–tail Hilbert unitary and the matrix-unit corner calculation in Section 10.2 below. Compression by the tail reference vector, followed by amplification with the tail identity, constructs the normal unital completely positive expectation

It is $A_n$-bimodular, preserves $\phi$, and commutes with $\sigma_t$. The following construction proves these assertions on the represented closure.

Faithful matrix vectors and concrete local slices

Here the slice map in (8) can be constructed directly. Write the product Hilbert space as $H_n\otimes H_{>n}$, and let $V_n\eta=\eta\otimes\Omega_{>n}$. On $B(H)$ put

Compression is normal: its vector coefficients are vector coefficients of $x$. Amplification is normal as well. For an elementary tensor coefficient, this follows by expanding its two vectors in finite sums of elementary tensor vectors; the uniform operator bound then gives the conclusion for all vector coefficients. Both maps are unital and completely positive, since their matrix amplifications preserve positivity by compression and by tensoring with the identity. Their composition is contractive and is bimodular over $B(H_n)\otimes1$.

On a finite left product, compression replaces each tail leg by its reference-vector value. To pass to the represented closure, choose an orthonormal basis $(f_u)$ of the finite-dimensional $H_n$. The amplified compression is the finite sum

It is therefore weak-operator continuous on every net, with no assumption that the approximating net is uniformly bounded. The finite-dimensional subspace $A_n\subset B(H_n)$ is weak-operator closed. The bicommutant closure of the local left algebra consequently maps into $A_n$, and compression fixes $A_n$. This restriction is precisely $E_n$. It preserves $\phi$, because $V_n\Omega_n=\Omega$, and commutes with the coordinate action because its implementing unitaries fix the reference vectors. For normality on arbitrary ultraweak nets, each coefficient is a vector functional and the displayed sum is finite; this also gives the exact all-net argument used in Section 10.3.

For completeness, the cyclicity used in Lemma 2.1 needs only finite matrix algebra. In one Hilbert--Schmidt coordinate, the maps $a\mapsto a\rho^{1/2}$ and $a\mapsto\rho^{1/2}a$ are bijective because $\rho$ is invertible. Their finite tensor products fill each finite-support Hilbert space. Consequently finite left and finite right products applied to $\Omega$ are dense in the pointed completion. Every right product commutes with the left products and their strong closure. An element of $M$ annihilating $\Omega$ therefore annihilates that dense right-product orbit and is zero. The vector state is normal, and $\phi(x^*x)=0$ now implies $x=0$, so it is faithful. The orthogonal-projection and dense-orbit argument in Lemma 2.1 consequently requires no modular theorem.

Lemma 2.1. For every $x\in M$, $E_n(x)\to x$ strongly*. Equivalently, the bounded sequence converges strongly both for $x$ and for $x^*$.

Proof. Let $Q_n$ project onto $A_n\Omega$. For $a\in A_n$, state preservation and bimodularity give

Since $E_n(x)\Omega\in A_n\Omega$, this proves

The subspaces increase and their union is dense, so $Q_n\to1$ strongly. Thus the difference in (9) tends to zero, as does the corresponding difference for $x^*$.

For every finite-support vector $\eta\in H_k$ and $n\ge k$, the same compression gives the stronger identity $E_n(x)\eta=Q_nx\eta$. Indeed, the tail of $\eta$ is the reference vector and $Q_n$ is precisely the projection onto $H_n\otimes\Omega_{>n}$. Thus convergence holds on all finite-support vectors. The differences have norm at most $2\|x\|$, so their dense union extends convergence to every vector. Apply compression separately to $x^*$, using $E_n(x)^*=E_n(x^*)$, to obtain the strong* assertion. Equation (9) also gives the commutant-orbit proof requested in Exercise 2. $\square$

Let $D_n$ be the diagonal algebra of $A_n$, and put

The product diagonal is identified with

The identification maps occupation projections to coordinate indicator functions. Here is its passage to the full bounded measurable algebra. Use the completed Bernoulli measure constructed in Section 3. Cylinder inner products agree exactly with the inner products of local diagonal vectors; the cylinder-density proof there therefore gives a unitary $\overline{D\Omega}\to L^2(X,\nu)$. To lift a bounded measurable $h$, average it over the cells $C_s$ of the first $n$ coordinates:

Every denominator is positive, $\|h_n\|_\infty\le\|h\|_\infty$, and these are the increasing finite-dimensional orthogonal projections of $h$ in $L^2$. Cylinder density proves $h_n\to h$ in $L^2$. Let $d_n\in D_n$ be the corresponding cylinder operator. Both $d_n\Omega$ and $d_n^*\Omega$ are Cauchy. For every finite right product $y'$, $\|(d_n-d_m)y'\Omega\|\le\|y'\|\,\|(d_n-d_m)\Omega\|$. The uniform norm bound extends this to every vector, and the same argument applies to adjoints. Thus $d_n$ converges strongly* to an element $d\in D$. On each cylinder vector its restriction is multiplication by $h$, and density gives the full multiplication operator.

Conversely, the finite coefficient formula for $E_n$ puts $E_n(d)\in D_n$ for $d\in D$: it holds on local diagonal products and passes to their weak-operator closure. The corresponding functions have uniform bound $\|d\|$, and Lemma 2.1 gives their $L^2$ limit $h$. This limit is essentially bounded with the same bound. Indeed, choose a subsequence with $L^2$ errors at most $2^{-2j}$; the sets where the error exceeds $2^{-j}$ have summable measures, and the union of each tail has measure tending to zero. Outside the resulting null set the subsequence converges pointwise. The restriction of $d$ to $\overline{D\Omega}$ is therefore multiplication by $h$. It is injective because $\Omega$ separates $M$. The lift above proves surjectivity, and restriction proves preservation of products and adjoints.

This is a normal identification in both directions. Restriction to the reducing subspace $\overline{D\Omega}$ is normal, since every compressed vector coefficient is a vector coefficient on $H$. Positivity is equivalent on the two sides: a positive $h$ has positive averages and a positive strong limit, and a positive $d$ has a positive multiplication restriction. The inverse order isomorphism consequently preserves least upper bounds of arbitrary bounded increasing positive nets. The concrete normality criterion, proved in Section 18 of the CAR lesson using positive functionals and summable vector coefficients, makes the inverse ultraweakly continuous on the entire algebra. Thus (10) is an identification of completed normal operator algebras, not merely of cylinder moments or an algebraic GNS subspace.

Lemma 2.2. $D$ is a maximal abelian subalgebra of $M$.

Proof. The local diagonals commute, so $D$ is abelian. If $x\in M$ commutes with $D$, then $E_n(x)$ commutes with $D_n$, by $A_n$-bimodularity. In $A_n$, the commutant of its full diagonal is just $D_n$: commuting with each diagonal matrix unit kills every off-diagonal entry. Thus $E_n(x)\in D_n$. Lemma 2.1 and strong closedness put $x\in D$. $\square$

3. The product-state centralizer is a factor

The centralizer is the fixed algebra

The restriction of $\phi$ to $N$ is a faithful normal tracial state by the following finite-density argument. If $x\in N$, equivariance gives $E_n(x)\in A_n^\sigma$, so $E_n(x)$ commutes with $\rho^{\otimes n}$. For every $y\in M$, bimodularity and state preservation reduce both $\phi(E_n(x)y)$ and $\phi(yE_n(x))$ to the finite matrices $E_n(x),E_n(y)$. Commutation with the density and cyclicity of the finite matrix trace make them equal. Lemma 2.1 passes the equality to $\phi(xy)=\phi(yx)$. Conversely, if this equality holds for every $y$, testing local matrix units makes $E_n(x)$ commute with $\rho^{\otimes n}$; equivariance and strong* convergence put $x\in N$. This establishes the centralizer identity itself on the completed representation. On $N$ it is tracial, and an isometry $v\in N$ has $\phi(1-vv^*)=0$, so faithfulness makes it unitary. Hence $N$ is finite. The diagonal is fixed by each coordinate density, so $D\subset N$.

For every finite permutation $\pi$ of the coordinate set, there is a unitary $v_\pi$ in a sufficiently large $A_n$ that permutes its tensor legs. The identical tensor densities satisfy

hence $v_\pi\in N$. On $D$, conjugation permutes the Bernoulli coordinates.

Direct diagonal density and permutation invariance

There is also an operator proof of the density and independence used in Lemma 3.1. Normality first shows that $E_n(D)\subset D_n$, by checking this on local diagonal products. Lemma 2.1 gives

Thus local diagonal elements are dense in the state $L^2$ norm, directly in the pointed representation.

Let $z\in D$ be invariant under finite permutations, and let $f$ be local diagonal. Choose local diagonal $g$ with $\|z-g\|_{2,\phi}<\varepsilon$. A finite permutation moves the support of $f$ disjointly from that of $g$; call the image $f_\pi$. The identical one-leg states imply that permutations preserve $\phi$. Invariance gives $\phi(zf)=\phi(zf_\pi)$, and disjoint finite tensor supports give exactly $\phi(gf_\pi)=\phi(g)\phi(f)$. Cauchy--Schwarz in the abelian state algebra gives

Let $\varepsilon$ tend to zero. Applying the equality to adjoints of all local $f$, and then using their state $L^2$ density, gives $(z-\phi(z)1)\Omega=0$. Faithfulness implies $z=\phi(z)1$. Under (10) this is the conclusion of Lemma 3.1. The calculation uses finite tensor independence, rather than an imported zero-one theorem.

Bernoulli measure and finite-permutation invariance

The following construction explains the measure and approximation steps used in Lemma 3.1. The domain is the countable product $X=\{0,1\}^{\mathbb N}$, with the identical Bernoulli probability measure in (10); the functions are complex elements of $L^\infty(X,\nu)$, and invariance means equality modulo null sets under every permutation fixing all but finitely many coordinates. The conclusion is constancy almost everywhere. We retain $0<p<1$, $p\ne\tfrac12$. The argument uses scalar measure and finite-coordinate independence directly.

We begin with ZFC with countable choice, completeness of the real and complex scalars, elementary finite sums/products and set operations, and the definitions of a sigma algebra, a measurable function and countable additivity. The construction proceeds from cylinders to the completed probability measure, scalar integration, $L^2$ approximation, and permutation invariance. The normal operator identification $D\cong L^\infty(X,\nu)$ in the other portion of (10) is separate from this scalar function argument.

The first four subsections construct and complete the measure. Integration and cylinder density then justify the error estimate. Finite independence and the permutation action give the final constancy argument.

Binary cylinders and function classes

Put $a=1-p$, $b=p$, so $a,b>0$ and $a+b=1$. For a finite set $F\subset\mathbb N$ and a binary assignment $\alpha:F\to\{0,1\}$, write

Empty products are one and $C(\varnothing,\varnothing)=X$. A cylinder function is any complex function determined by a finite coordinate set; such a function has finite range. Let $\mathcal A_n$ be the finite algebra of unions of the $2^n$ first-$n$ cylinder cells, and let $\mathcal A=\bigcup_n\mathcal A_n$, including $\varnothing,X$. It is an algebra, because any finite collection of its sets is contained in one $\mathcal A_n$.

Write $\mathcal B=\sigma(\mathcal A)$. This is the coordinate product sigma algebra: every finite-coordinate cylinder is a union of first-$n$ cells for sufficiently large $n$, and the first-$n$ cells are finite intersections of coordinate inverse images. It is also the Borel sigma algebra of the usual product topology. Indeed the prefix cylinders form a countable base; every open set is a union of a subcollection of this countable base, and each cylinder is open and closed. No uncountable-product theorem is involved.

The finite-weight construction, outer-measure extension, and completion construct the probability $\nu_0$ on $\mathcal B$ and its complete extension $\nu$ on $\overline{\mathcal B}$, with

The last notation means precisely the unique countably additive measure with these cylinder values, together with completion when completed function classes are used. Passing from $\mathcal B$ to its completion changes no $L^2$ or $L^\infty$ equivalence class: completion and set approximation and cylinder density and completeness prove the necessary representative statement. Thus both usual readings of original (10) give the same lemma.

Let $\Gamma$ be the group of permutations $\pi$ of $\mathbb N$ with finite support. We fix the explicit coordinate and function conventions

The permutation construction proves that these are well-defined measure-preserving actions, also on completed equivalence classes. The target is: if $z\in L^\infty(X,\nu)$ and $\alpha_\pi z=z$ for every $\pi\in\Gamma$, then $z=c$ almost everywhere for the scalar $c=\int_X z\,d\nu$. Pointwise invariance of a chosen representative is not assumed. Since the original invariance quantifies over every finite permutation, replacing each permutation by its inverse gives exactly the same hypothesis under the alternative coordinate convention.

Finite Bernoulli weights and cylinder covers

For a first-$n$ cell with string $\beta\in\{0,1\}^n$, set $m(C_n(\beta))=a^{n-|\beta|}b^{|\beta|}$, and for a union of these cells sum their weights. Refinement from level $n$ to $n+1$ is consistent, since

Thus $m$ is defined independently of the level on all of $\mathcal A$. Its finite partition formula proves nonnegativity, finite additivity, monotonicity and $m(X)=(a+b)^n=1$. In particular $m(A\cup B)\le m(A)+m(B)$. Refining a finite-coordinate cylinder to a head partition and summing the unspecified binary digits proves $m(C(F,\alpha))=w(F,\alpha)$, because each missing digit contributes $a+b=1$.

We prove the finite-cover property needed for countable additivity and outer measure, rather than invoking compactness. Suppose $A\in\mathcal A$ is covered by a family of sets from $\mathcal A$ but no finite subfamily covers it. Choose a head level expressing $A$ as finitely many prefix cells. One of these cells has no finite subcover; otherwise the finitely many finite subcovers combine. Split that cell into its two children. Again one child has no finite subcover. Recursion gives nested prefix cells of lengths tending to infinity, each with no finite subcover. Their compatible digits define a single $x\in A$. Some member $B$ of the given cover contains $x$. Since $B$ is a union of cells at one finite level, the prefix cell of $x$ at every sufficiently large level is contained in $B$. A cell in the recursive family is then covered by this one member, a contradiction. This proves the finite-cover property for every such cover of $A$.

Consequently, whenever $A\subseteq\bigcup_{k\ge1}A_k$ with $A,A_k\in\mathcal A$, a finite subcover and finite additivity give

If the $A_k$ are disjoint and their union itself belongs to $\mathcal A$, a finite subcover covers that union. Every other $A_k$ must then be empty, by disjointness. The finite additivity formula therefore proves countable additivity on exactly those unions required of a premeasure. No product-measure existence or general compactness theorem is hidden in this step.

Outer measure and countable additivity

For every subset $E\subseteq X$, define

There is always a cover, using $X$, so $0\le\nu^*(E)\le1$. The empty set has value zero and the definition gives monotonicity. For countable subadditivity, the assertion is immediate if $\sum_j\nu^*(E_j)=\infty$. Otherwise, for a prescribed $\varepsilon>0$, choose for each $j\ge1$ a cylinder-algebra cover of $E_j$ with total weight less than $\nu^*(E_j)+\varepsilon2^{-j}$. Their countable union is a cover of $\bigcup_jE_j$. Infimizing and then letting $\varepsilon\downarrow0$ gives

For $A\in\mathcal A$, the one-set cover gives $\nu^*(A)\le m(A)$; (FE5) applied to every countable cover gives the reverse inequality. Hence $\nu^*|_{\mathcal A}=m$.

Call $E$ measurable for this outer measure if, for every subset $T\subseteq X$,

Let $\mathcal C$ be the collection of these sets. Every $A\in\mathcal A$ satisfies (FE8): split each member $A_k$ of any cover of $T$ into $A_k\cap A$ and $A_k\setminus A$. Both parts are in $\mathcal A$, with weights summing to $m(A_k)$. Thus the right side of (FE8) is at most the cost of any cover of $T$, hence at most $\nu^*(T)$; subadditivity gives the opposite inequality.

We supply the whole sigma-algebra and measure argument. The empty set belongs to $\mathcal C$, and complements preserve (FE8). If $E,F\in\mathcal C$, successively split $T$ by $E$, and then its two parts by $F$. The resulting four-term equality implies the test for $E\cup F$: the sum of the first three terms is at least $\nu^*(T\cap(E\cup F))$ by subadditivity, while the fourth is $\nu^*(T\setminus(E\cup F))$. This supplies the lower bound in (FE8); its upper bound always follows from subadditivity. Thus $\mathcal C$ is an algebra.

For disjoint $E_j\in\mathcal C$, write $E=\bigcup_{j\ge1}E_j$. Repeated finite splitting gives, for every $n$,

Let $n\to\infty$. Subadditivity bounds $\nu^*(T\cap E)$ by the resulting sum. This proves the lower bound in (FE8) for $E$, and the opposite bound is automatic. Therefore $E\in\mathcal C$. An arbitrary countable union can be replaced by the disjoint sets $E_1,E_2\setminus E_1,\ldots$, using the algebra operations just proved. Hence $\mathcal C$ is a sigma algebra containing $\mathcal B$.

For disjoint measurable $E_j$, take $T=E$ in the finite splitting equality. It gives $\nu^*(E)\ge\sum_{j\le n}\nu^*(E_j)$. Let $n\to\infty$ and combine with (FE7). Thus

The restriction to $\mathcal C$, and in particular to $\mathcal B$, is a countably additive probability. If $\nu^*(Z)=0$, every subset of $Z$ has outer measure zero. For such a subset the right side of (FE8) is at most $\nu^*(T)$ by monotonicity, and at least it by subadditivity. It is measurable. The measure on $\mathcal C$ is therefore complete.

Completion, uniqueness and set approximation

Write $\nu=\nu^*|_{\mathcal C}$ provisionally. Countable additivity implies the elementary continuity statements used here. If $E_n\uparrow E$, write $E=E_1\sqcup\bigsqcup_{n\ge2}(E_n\setminus E_{n-1})$. The partial sums are $\nu(E_n)$, so $\nu(E_n)\uparrow\nu(E)$. If $E_n\downarrow E$, take complements in the finite measure $X$, obtaining $\nu(E_n)\downarrow\nu(E)$. These proofs apply throughout $\mathcal C$.

We identify $\mathcal C$ exactly with the completion of $\mathcal B$. For $E\in\mathcal C$, choose for every $n$ a cylinder-algebra cover of total weight less than $\nu^*(E)+2^{-n}$, and let $U_n\in\mathcal B$ be its union. Then $E\subseteq U_n$ and countable subadditivity plus measurable subtraction give

Let $B=\bigcap_nU_n\in\mathcal B$. Then $E\subseteq B$ and $\nu^*(B\setminus E)=0$. Every outer-null set $Z$ is contained in a Borel null set: choose covers with total weights less than $2^{-n}$, take the union of each cover and then their intersection. The resulting Borel set contains $Z$ and has measure at most $2^{-n}$ for every $n$. Hence $E$ differs from $B$ by a subset of a Borel null set. Conversely every such completed set belongs to $\mathcal C$, since Borel sets and all subsets of null sets belong to it. This proves $\mathcal C=\overline{\mathcal B}$, and we now write $\nu=\nu^*|_{\overline{\mathcal B}}$.

For uniqueness on $\mathcal B$, let $\lambda$ be any probability agreeing with $m$ on $\mathcal A$. For $B\in\mathcal B$, each cover in (FE6) gives $\lambda(B)\le\sum_km(A_k)$, so $\lambda(B)\le\nu_0(B)$. Apply this to $X\setminus B$, and use both probabilities' value one on $X$, to obtain the reverse inequality. Thus $\lambda=\nu_0$. Its complete extension is also unique, because a completed set differs from a Borel set by a subset of a Borel null set, which every complete extension measures zero. This proves the product measure assertion (FE2), without invoking an extension, uniqueness or infinite-product theorem.

For every $E\in\overline{\mathcal B}$ and $\delta>0$, choose a cover with union $U$ and total cost less than $\nu(E)+\delta/2$. Then $\nu(U\setminus E)<\delta/2$. Let $A_N$ be the union of its first $N$ members, so $A_N\in\mathcal A$ and $A_N\uparrow U$. By the continuity already proved, some $N$ has $\nu(U\setminus A_N)<\delta/2$. Consequently

This is a complete set-approximation theorem for the completed measure. No monotone-class or density theorem is assumed.

Scalar integration and square norms

For a complex simple function $s=\sum_{\ell=1}^r c_\ell1_{E_\ell}$ on a finite measurable partition of $X$, define

Common finite refinement shows independence of the representation, complex linearity and positivity. It also gives $|I(s)-I(t)|\le\sup_X|s-t|$ for simple $s,t$. A bounded measurable real function admits uniform simple approximations by subdividing its bounded range into intervals of length $2^{-n}$; for a complex function apply this to its real and imaginary parts. Such approximations are measurable because their finitely many level sets are measurable. Completeness of the complex scalars makes $I(s_n)$ converge, and the preceding estimate makes the limit independent of the approximating sequence. This defines $I(f)$ for all bounded measurable complex $f$, with positivity, linearity and the uniform estimate retained.

For a nonnegative measurable $u$, possibly taking $+\infty$, define $I(u)$ to be the supremum of $I(s)$ over nonnegative simple $s\le u$. For bounded nonnegative $u$ this equals its preceding uniform integral, by lower dyadic simple approximation. We prove monotone convergence directly. If $u_n\uparrow u$, monotonicity gives $\sup_nI(u_n)\le I(u)$. Fix a nonnegative simple $s\le u$ and $0<c<1$. The measurable sets $F_n=\{u_n\ge cs\}$ increase and exhaust each cell where $s$ has positive value. The continuity of measure from completion and set approximation therefore gives $I(s1_{F_n})\to I(s)$. Since $u_n\ge cs1_{F_n}$,

Zero-valued cells do not affect this argument.

The explicit bounded simple functions

are increasing: finer binary subdivision increases the lower approximation, and increasing the cutoff cannot decrease it. Applying (FE14) to $q_n(u)+q_n(v)$ proves $I(u+v)=I(u)+I(v)$ for nonnegative functions; their finite simple integrals are already additive. The same lower-approximation argument proves positive homogeneity. These formulas hold with nonnegative infinite values. If $u_n\ge0$, then $v_n=\inf_{k\ge n}u_k\uparrow\liminf_ku_k$ is measurable, and $I(v_n)\le\inf_{k\ge n}I(u_k)$. Monotone convergence proves

Measurability of countable suprema and infima follows from their threshold sets being countable unions or intersections.

For real $f$ with $I(|f|)<\infty$, put $I(f)=I(f^+)-I(f^-)$; integrate complex functions by real and imaginary parts. Additivity for integrable real functions follows by integrating the pointwise identity

All its terms are integrable because $|f+g|\le|f|+|g|$; nonnegative additivity was proved above. Homogeneity and complex linearity follow as well. Rotating the value of $I(f)$ by a unit scalar and using $\operatorname{Re}(\zeta f)\le|f|$ proves $|I(f)|\le I(|f|)$. We henceforth use the ordinary notation $\int f\,d\nu=I(f)$.

Every finite-$L^2$-norm function is finite almost everywhere: if $I(|u|^2)<\infty$, the set where $|u|=+\infty$ has measure at most $I(|u|^2)/R^2$ for every $R>0$. It has measure zero. For complex measurable $u,v$ with finite square integrals, the scalar inequality

shows that $uv$ is integrable. Infimizing its integrated bound in $t$ gives

When one squared norm is zero, let $t\to0$ or $t\to\infty$ in the finite bound; thus no division by zero is used. Taking $v=1$ also proves integrability of every $L^2$ function. Expanding $|u+v|^2$, using the now justified complex integral and (FE18), proves the triangle inequality

Linearity of the measurable operations used here is literal pointwise arithmetic. Their measurability follows from continuity of scalar arithmetic and the countable rectangular bases in finite-dimensional real or complex scalar spaces: an inverse image of an open set is a countable union of measurable coordinate inverse images. For a convergent sequence of measurable real functions, the threshold set $\{\lim_n u_n<r\}$ is, on its convergence set,

Indeed a limit below $r$ admits a rational $q$ strictly between the limit and $r$, whereas eventual values below $q<r$ force the limit to be at most $q<r$. The set of convergence of a real or complex scalar sequence is itself measurable by completeness and the Cauchy formula