Normalized phase volume and almost-everywhere assertions

Assume countable choice. Let $V$ be a nonempty finite set, let $n=\mathrm{\mid }V\mathrm{\mid }$, and identify ${\mathbb{R}}^V$ with ${\mathbb{R}}^n$ by any ordering of $V$. Put $T_V={\mathbb{R}}^V\mathrm{/}(2\pi \mathbb{Z}{)}^V$, with its quotient topology, and write $q$ for the quotient map. Put $D=(-\pi ,\pi {]}^V$. A subset $E\subseteq T_V$ is phase-measurable exactly when $q^{-1}(E)\cap D$ is Lebesgue measurable; its normalized phase volume is $\mu (E)=(2\pi {)}^{-n}{\lambda }_n(q^{-1}(E)\cap D)$. This is the complete probability measure obtained by pushing normalized Lebesgue measure on $D$ through $q$. Write ${\mathcal{M}}_V$ for the sigma-algebra of phase-measurable sets. Phase measurability and normalized phase volume are independent of the ordering of $V$.

A set is phase-null when it is phase-measurable and has volume zero. A statement holds for almost every initial phase when it holds outside a phase-null set. A uniformly sampled phase configuration is a random element $X:(\mathrm{\Omega },\mathcal{F},\mathbb{P})\to (T_V,{\mathcal{M}}_V)$ on a probability space, whose law is $\mu$: for every $E\in {\mathcal{M}}_V$, $\mathbb{P}(X^{-1}(E))=\mu (E)$. An angular chart is the inverse of $q$ restricted to a product of open intervals, each of length less than $2\pi$; its domain is the image of that product. A subset of a chart domain is phase-null exactly when its set of chart coordinates is Lebesgue-null. Every nonempty open subset of $T_V$ has positive phase volume. Phase volume is a measure on phase space; a separately specified probability law for a random graph is a different measure.

The quotient topology of a surjection, quotient maps, saturated sets, and the quotient of a space by an equivalence relation with its canonical projection

The product set ∏_i ∈ I Xᵢ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space

The Axiom of Countable Choice (AC_ω)

Lebesgue measurable sets, the family L(ℝⁿ), and the restricted set function λₙ

Probability measures and probability spaces

Random elements and real random variables

Law or distribution of a random element

The Borel sigma-algebra of a topological space

Assuming countable choice, every Borel subset of ℝⁿ is Lebesgue measurable

Lebesgue measure on ℝⁿ is invariant under every orthogonal linear map

Sigma-algebras

Measures on sigma-algebras

Measurable spaces and measurable sets

Complete measure spaces

Proof technique: direct.

1.1 Each coordinate has a unique representative in $(-\pi ,\pi ]$ modulo $2\pi \mathbb{Z}$, so $q{\mathrm{\mid }}_D$ is a bijection. Inverse images under this bijection preserve complements and countable unions. Consequently the displayed measurable sets form a sigma-algebra, and their assigned values form a measure by countable additivity of Lebesgue measure. The box formula gives ${\lambda }_n(D)=(2\pi {)}^n$, so the total mass is one. If a phase-measurable set has zero volume, the inverse image in $D$ of any subset is a subset of a Lebesgue-null set and hence is measurable and null; thus this pushforward measure is complete. [given, F1, F8]

1.2 Let $a,b:{\mathbb{R}}^V\to {\mathbb{R}}^n$ be the coordinate identifications from two orderings of $V$, and put $P=b\circ a^{-1}$. This is a linear bijection permuting coordinates, with inverse given by the reverse permutation. Reindexing the finite sum gives ${\sum }_{i=1}^n(Px{)}_i(Py{)}_i={\sum }_{i=1}^n{x}_i{y}_i$, so $P$ is orthogonal. For every $E\subseteq T_V$ it carries $a(D)$ to $b(D)$ and $a(q^{-1}(E)\cap D)$ to $b(q^{-1}(E)\cap D)$. Apply orthogonal invariance to $P$ and $P^{-1}$: one representative set is Lebesgue measurable exactly when the other is, and their measures agree. The factor $(2\pi {)}^{-n}$ is the same for both orderings, proving independence of phase measurability and volume. Coordinate permutations also carry open boxes to open boxes, so the product and quotient topologies used here do not depend on the ordering. [given, F3, F6]

2.1 By step 1.1, $(T_V,{\mathcal{M}}_V,\mu )$ is a probability space. Its identity map is measurable because the inverse image of $E\in {\mathcal{M}}_V$ is $E$ itself, and its law assigns $E$ the value $\mu (E)$. It is therefore a uniformly sampled phase configuration in the stated sense, so this formulation has a realization on a probability space. [step 1.1, F7, F8]

2.2 Let $U$ be a product of intervals of length less than $2\pi$. Distinct points of $U$ cannot differ by a nonzero vector in $(2\pi \mathbb{Z}{)}^V$, so $q{\mathrm{\mid }}_U$ is injective. It is an open map onto its image: the inverse image under $q$ of the image of an open subset of $U$ is the union of its lattice translates, which is open. Thus its inverse is an angular chart. For $A\subseteq U$, the set $q^{-1}(q(A))\cap D$ is the countable union over $k\in {\mathbb{Z}}^V$ of $(A+2\pi k)\cap D$. Translation invariance and countable subadditivity show that if $A$ is Lebesgue-null then this inverse image is null. Conversely, $A$ is contained in the countable union of translates of $q^{-1}(q(A))\cap D$; if the latter is null, then so is $A$. Completeness makes each null set in this argument measurable, proving the chart equivalence. [step 1.1, F1, F2, F3, F8, F9]

3.1 If $O\subseteq T_V$ is nonempty and open, its inverse image is nonempty and open. It contains a nonempty product $B$ of bounded open intervals, each shorter than $2\pi$. By the box formula $B$ has positive Lebesgue measure. If $O$ had zero phase volume, the chart equivalence would make $B$ null, a contradiction. Also $q^{-1}(O)$ is open, and $D$ is Borel, since it is the finite intersection of the open coordinate conditions $-\pi <x_v$ and the closed coordinate conditions $x_v\le \pi$. Thus $q^{-1}(O)\cap D$ is Borel and is Lebesgue measurable under countable choice, proving that $O$ is phase-measurable. Since ${\mathcal{M}}_V$ is a sigma-algebra containing the open sets, it contains the Borel sigma-algebra of $T_V$. [step 1.1, step 2.2, F1, F3, F5]

4.1 Scaling representatives by $1\mathrm{/}(2\pi )$ induces a bijection from $T_V$ to ${\mathbb{R}}^n\mathrm{/}{\mathbb{Z}}^n$, and $D$ becomes $(-1\mathrm{/}2,1\mathrm{/}2{]}^n$. Scaling and inverse scaling carry Euclidean open sets to open sets, so the induced quotient bijection is a homeomorphism. Its inverse images preserve complements and countable unions, so it and its inverse preserve Borel sets. For a Borel set, the dilation formula turns the normalized volume in the definition into the period-one volume computed on this translated half-open cube. Hence the construction has the stated $2\pi$ normalization and agrees on Borel sets with the supplied finite-torus measure. The completed pushforward in step 1.1 supplies the phase-null and almost-everywhere conventions without imposing a law on any graph. [step 1.1, step 3.1, F2, F3, F4, F5] ∎