Proof of Tensor Powers and One-Particle Marginals: Particle Laws, Product Integrals, Push-Forwards, Moments, Product Maps and Diagonal Shifts

Each result cited is universally quantified over the data in its own statement. For $n\in\mathbb{N}$ write $\mathfrak{p}^{(n)}_{k}$ ($k\in[n]$) for the block maps of $\mathbb{R}^{qn}$ of Particle Blocks of the Configuration Space: Block Maps, Configurations, Product Maps and Diagonal Points §blocks with $n$ in place of $N$, so $\mathfrak{p}_{k}=\mathfrak{p}^{(N)}_{k}$, and $\hat{\mathfrak{p}}_{k}$ for the block maps of $\mathbb{R}^{pN}$. Tensor powers $\rho^{\otimes n}$ for every $n\in\mathbb{N}$ are those of The Tensor Power of a Probability Measure on Euclidean Space §tensor, and by The One-Particle Marginal of a Probability Measure on the Configuration Space §marginal the one-particle marginal $P^{[1]}$ is the measure $A_{P}$ of Existence and Uniqueness of Tensor Powers, and the Average of the Block Marginals §average. The product of Finite Product Notation satisfies the recursion defining the product of Finite Product Notation in a Field for the field $\mathbb{R}$, so the two coincide by the uniqueness in Existence and Uniqueness of Iterates of a Binary Operation, and Properties of Finite Products applies to it. Finite sums in $[0,\infty]$ are the iterates of Existence and Uniqueness of Iterates of a Binary Operation for the addition of $[0,\infty]$ fixed in Measure, Measure Space, and Probability Measure; by the uniqueness there they satisfy $\sum_{k=1}^{1}c_{k}=c_{1}$ and $\sum_{k=1}^{n+1}c_{k}=\sum_{k=1}^{n}c_{k}+c_{n+1}$, and agree with finite sums of real numbers when all summands are real. Natural numbers are read in $\mathbb{R}$ through the canonical map, positive and invertible by claim 3 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field (The Real Numbers: Standing Notation and Background §numbers).

Step 0 (Preliminaries). (i) $q\cdot1=q$ by Natural Numbers, and $b(1,i)=(1-1)q+i=i$ for $i\in[q]$ by the axioms of Field, so $\mathfrak{p}^{(1)}_{1}(x)=x$ for $x\in\mathbb{R}^{q}$ by Particle Blocks of the Configuration Space: Block Maps, Configurations, Product Maps and Diagonal Points §blocks. As $[1]=\{1\}$ (claim 2 of Basic Properties of Initial Segments of the Natural Numbers) and $\prod_{k=1}^{1}\rho(B_{k})=\rho(B_{1})$, the measure $\rho$ satisfies the defining identity of $\rho^{\otimes1}$, so $\rho^{\otimes1}=\rho$ by the uniqueness in Existence and Uniqueness of Tensor Powers, and the Average of the Block Marginals §tensor.

(ii) Let $n\in\mathbb{N}$. By Natural Numbers, $q(n+1)=qn+q$, and we write $\iota=\iota^{qn,q}$, $\mathrm{pr}_{1}=\mathrm{pr}^{qn,q}_{1}$, $\mathrm{pr}_{2}=\mathrm{pr}^{qn,q}_{2}$ (Probability Measures on Euclidean Space and Random Vectors: Standing Notation §pairs). For $z\in\mathbb{R}^{qn+q}$ we have $z=\iota(\mathrm{pr}_{1}(z),\mathrm{pr}_{2}(z))$ by Pairs of Euclidean Points: Coordinate Projections, Pairings, the Product Measure on a Euclidean Space, Borel Norm Functions and Finite Sets §projections, hence by Particle Blocks: Linearity, Splitting of Inner Products, Product Maps and Diagonal Shifts §concatenation

Moreover $\rho^{\otimes(n+1)}=\rho^{\otimes n}\boxtimes\rho$: for $B_{1},\dots,B_{n+1}\in\mathcal{B}(\mathbb{R}^{q})$ let $A=\bigcap_{k=1}^{n}(\mathfrak{p}^{(n)}_{k})^{-1}(B_{k})\in\mathcal{B}(\mathbb{R}^{qn})$ (Particle Blocks: Linearity, Splitting of Inner Products, Product Maps and Diagonal Shifts §linear); since $[n+1]=[n]\cup\{n+1\}$ (claim 3 of Basic Properties of Initial Segments of the Natural Numbers), the display gives $\bigcap_{k=1}^{n+1}(\mathfrak{p}^{(n+1)}_{k})^{-1}(B_{k})=\mathrm{pr}_{1}^{-1}(A)\cap\mathrm{pr}_{2}^{-1}(B_{n+1})=\iota(A\times B_{n+1})$, whose $\rho^{\otimes n}\boxtimes\rho$-measure is $\rho^{\otimes n}(A)\rho(B_{n+1})=\bigl(\prod_{k=1}^{n}\rho(B_{k})\bigr)\rho(B_{n+1})=\prod_{k=1}^{n+1}\rho(B_{k})$ by Pairs of Euclidean Points: Coordinate Projections, Pairings, the Product Measure on a Euclidean Space, Borel Norm Functions and Finite Sets §product, The Tensor Power of a Probability Measure on Euclidean Space §tensor and Finite Product Notation ($2\le n+1$ by claims 4 and 6 of Properties of the Order on the Natural Numbers); the uniqueness in Existence and Uniqueness of Tensor Powers, and the Average of the Block Marginals §tensor gives the claim.

(iii) With $n=1$, (i) and (ii) give, on $\mathbb{R}^{q\cdot2}=\mathbb{R}^{q+q}$, $\mathfrak{p}^{(2)}_{1}=\mathrm{pr}^{q,q}_{1}$, $\mathfrak{p}^{(2)}_{2}=\mathrm{pr}^{q,q}_{2}$ and $\rho^{\otimes2}=\rho\boxtimes\rho$.

(iv) For real $c$ and $n\in\mathbb{N}$, $\sum_{k=1}^{n}c=nc$: the set of $n$ for which this holds for every $c$ contains $1$ and contains $n+1$ with $n$, by claim 1 of Properties of Finite Sums, claim 1 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field and the axioms of Field, so it is $\mathbb{N}$ by Principle of Induction for the Natural Numbers.

(v) Let $(X,\mathcal{F},\mu)$ be a measure space. The set $V$ of those $n\in\mathbb{N}$ such that for all measurable $g_{1},\dots,g_{n}:X\to[0,\infty)$ the function $\sum_{k=1}^{n}g_{k}$ is measurable with $\int_{X}\sum_{k=1}^{n}g_{k}\,d\mu=\sum_{k=1}^{n}\int_{X}g_{k}\,d\mu$ in $[0,\infty]$ contains $1$ by claim 1 of Properties of Finite Sums; and if $n\in V$, then $\sum_{k=1}^{n+1}g_{k}=\sum_{k=1}^{n}g_{k}+g_{n+1}$ pointwise (claim 1 of Properties of Finite Sums), with nonnegative values by claim 5 there, so claim 1 of Linearity and Monotonicity of the Lebesgue Integral and $n\in V$ give $n+1\in V$. Hence $V=\mathbb{N}$ by Principle of Induction for the Natural Numbers.

(vi) For $a\in[0,\infty]$: if $a=\infty$ then $N^{-1}a=\infty$ and $Na=\infty$ (Measure Spaces and the Lebesgue Integral: Standing Notation §extended); if $a$ is real, so are $N^{-1}a$ and $Na$, and $N(N^{-1}a)=a$ by Field. Hence $N(N^{-1}a)=a$, and $Na<\infty$, $a<\infty$, $N^{-1}a<\infty$ are equivalent.

Claim 1. Let $k\in[N]$ and $B\in\mathcal{B}(\mathbb{R}^{q})$. Put $B_{k}=B$ and $B_{l}=\mathbb{R}^{q}$ for $l\ne k$; as $\mathfrak{p}_{l}^{-1}(\mathbb{R}^{q})=\mathbb{R}^{qN}$, $\bigcap_{l=1}^{N}\mathfrak{p}_{l}^{-1}(B_{l})=\mathfrak{p}_{k}^{-1}(B)$, and $\rho(B_{l})=1$ for $l\ne k$. By Probability Measures on Euclidean Space and Random Vectors: Standing Notation §pushforward, The Tensor Power of a Probability Measure on Euclidean Space §tensor and claim 3 of Properties of Finite Products, $(\mathfrak{p}_{k})_{\#}\rho^{\otimes N}(B)=\rho^{\otimes N}(\mathfrak{p}_{k}^{-1}(B))=\prod_{l=1}^{N}\rho(B_{l})=\rho(B)$.

Let $k,l\in[N]$ with $k\ne l$, and $u=(\mathfrak{p}_{k},\mathfrak{p}_{l}):\mathbb{R}^{qN}\to\mathbb{R}^{q+q}$, Borel by Pairs of Euclidean Points: Coordinate Projections, Pairings, the Product Measure on a Euclidean Space, Borel Norm Functions and Finite Sets §pairing and Particle Blocks: Linearity, Splitting of Inner Products, Product Maps and Diagonal Shifts §linear, with $\mathrm{pr}^{q,q}_{1}\circ u=\mathfrak{p}_{k}$ and $\mathrm{pr}^{q,q}_{2}\circ u=\mathfrak{p}_{l}$ by Pairs of Euclidean Points: Coordinate Projections, Pairings, the Product Measure on a Euclidean Space, Borel Norm Functions and Finite Sets §projections. Let $Q=u_{\#}\rho^{\otimes N}\in\mathcal{P}(\mathbb{R}^{q+q})$ (Probability Measures on Euclidean Space and Random Vectors: Standing Notation §pushforward). For $B_{1},B_{2}\in\mathcal{B}(\mathbb{R}^{q})$, by Step 0(iii),

where $C_{k}=B_{1}$, $C_{l}=B_{2}$ and $C_{m}=\mathbb{R}^{q}$ otherwise. Let $a_{m}=\rho(B_{1})$ for $m=k$ and $a_{m}=1$ otherwise, and $b_{m}=\rho(B_{2})$ for $m=l$ and $b_{m}=1$ otherwise; then $\rho(C_{m})=a_{m}b_{m}$ for every $m\in[N]$. By The Tensor Power of a Probability Measure on Euclidean Space §tensor and claims 2 and 3 of Properties of Finite Products, $Q$ gives this set the value $\prod_{m=1}^{N}a_{m}b_{m}=\rho(B_{1})\rho(B_{2})=\prod_{m=1}^{2}\rho(B_{m})$. So $Q$ satisfies the defining identity of $\rho^{\otimes2}$, and $Q=\rho^{\otimes2}=\rho\boxtimes\rho$ by the uniqueness in Existence and Uniqueness of Tensor Powers, and the Average of the Block Marginals §tensor and Step 0(iii).

Claim 2. Let $T$ be the set of those $n\in\mathbb{N}$ such that for all bounded Borel $f_{1},\dots,f_{n}:\mathbb{R}^{q}\to\mathbb{R}$ the function $F_{n}=\prod_{k=1}^{n}f_{k}\circ\mathfrak{p}^{(n)}_{k}$ on $\mathbb{R}^{qn}$ is Borel and bounded and $\int F_{n}\,d\rho^{\otimes n}=\prod_{k=1}^{n}\int f_{k}\,d\rho$; the integrals exist by Probability Measures on Euclidean Space and Random Vectors: Standing Notation §measures. By Step 0(i), $F_{1}=f_{1}$ and $\rho^{\otimes1}=\rho$, so $1\in T$.

Let $n\in T$ and let $f_{1},\dots,f_{n+1}$ be bounded Borel. Let $G=\prod_{k=1}^{n}f_{k}\circ\mathfrak{p}^{(n)}_{k}$, which is Borel and bounded with $\int G\,d\rho^{\otimes n}=\prod_{k=1}^{n}\int f_{k}\,d\rho$ because $n\in T$ (claim 1 of Properties of Finite Products), and let $f=f_{n+1}$. By Finite Product Notation and Step 0(ii), $F_{n+1}(z)=G(\mathrm{pr}_{1}(z))\,f(\mathrm{pr}_{2}(z))$ for $z\in\mathbb{R}^{qn+q}$. It is Borel by Probability Measures on Euclidean Space and Random Vectors: Standing Notation §borel-maps, Pairs of Euclidean Points: Coordinate Projections, Pairings, the Product Measure on a Euclidean Space, Borel Norm Functions and Finite Sets §projections and claim 3 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions; and if $M_{G},M_{f}$ are bounds for $G,f$ (Bounded Real-Valued Function on a Set), then $|F_{n+1}(z)|=|G(\mathrm{pr}_{1}(z))|\,|f(\mathrm{pr}_{2}(z))|\le M_{G}M_{f}$ by claims 4 and 1 of Properties of the Absolute Value in an Ordered Field and claim 5 of Elementary Arithmetic in an Ordered Field. Let $\mu=\rho^{\otimes n}$. By Step 0(ii) and Pairs of Euclidean Points: Coordinate Projections, Pairings, the Product Measure on a Euclidean Space, Borel Norm Functions and Finite Sets §product, $\rho^{\otimes(n+1)}$ is the image measure of $\mu\otimes\rho$ under the measurable map $\iota$; as $F_{n+1}$ is integrable with respect to $\rho^{\otimes(n+1)}$ (Probability Measures on Euclidean Space and Random Vectors: Standing Notation §measures), claim 2 of Image Measures, Measures with Densities, and Change of Variables shows that $H=F_{n+1}\circ\iota$, given by $H(x,y)=G(x)f(y)$ (Pairs of Euclidean Points: Coordinate Projections, Pairings, the Product Measure on a Euclidean Space, Borel Norm Functions and Finite Sets §projections), is integrable with respect to $\mu\otimes\rho$ and $\int F_{n+1}\,d\rho^{\otimes(n+1)}=\int H\,d(\mu\otimes\rho)$. The measures $\mu,\rho$ are $\sigma$-finite (Measure, Measure Space, and Probability Measure, with every $X_{m}$ the whole space). By the Fubini part of Tonelli and Fubini Theorems there is $N_{0}\in\mathcal{B}(\mathbb{R}^{qn})$ with $\mu(N_{0})=0$ such that the function $\varphi$ equal to $\int f_{x}\,d\rho$ at $x\notin N_{0}$, where $f_{x}(y)=G(x)f(y)$, and to $0$ on $N_{0}$, is integrable with respect to $\mu$ and $\int H\,d(\mu\otimes\rho)=\int\varphi\,d\mu$. Put $c=\int f\,d\rho$. By claim 2 of Linearity and Monotonicity of the Lebesgue Integral, $\int f_{x}\,d\rho=G(x)c$ for every $x$, so $\varphi$ and $cG$ agree off $N_{0}$, that is almost everywhere (Null Set of a Measure, A Property Holding Almost Everywhere). As $G$ is integrable with respect to $\mu$ (Probability Measures on Euclidean Space and Random Vectors: Standing Notation §measures), The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere §comparison and claim 2 of Linearity and Monotonicity of the Lebesgue Integral give $\int\varphi\,d\mu=\int cG\,d\mu=c\int G\,d\mu$. Hence

by Field and Finite Product Notation, and $n+1\in T$. By Principle of Induction for the Natural Numbers, $T=\mathbb{N}$; $N\in T$ is claim 2.

Claim 3. The argument uses only $p\in\mathbb{N}$, so it holds with any natural number in place of $p$; Claim 7 uses it with $q$ in place of $p$. The map $h^{\oplus}$ is Borel and $\hat{\mathfrak{p}}_{k}\circ h^{\oplus}=h\circ\mathfrak{p}_{k}$ for $k\in[N]$ by Particle Blocks: Linearity, Splitting of Inner Products, Product Maps and Diagonal Shifts §product-map, so $(h^{\oplus})^{-1}(\hat{\mathfrak{p}}_{k}^{-1}(B))=\mathfrak{p}_{k}^{-1}(h^{-1}(B))$ for $B\in\mathcal{B}(\mathbb{R}^{p})$, with $h^{-1}(B)\in\mathcal{B}(\mathbb{R}^{q})$. For $B_{1},\dots,B_{N}\in\mathcal{B}(\mathbb{R}^{p})$, by Probability Measures on Euclidean Space and Random Vectors: Standing Notation §pushforward and The Tensor Power of a Probability Measure on Euclidean Space §tensor,

so, $(h^{\oplus})_{\#}\rho^{\otimes N}$ and $(h_{\#}\rho)^{\otimes N}$ both lying in $\mathcal{P}(\mathbb{R}^{pN})$, they are equal by the uniqueness in Existence and Uniqueness of Tensor Powers, and the Average of the Block Marginals §tensor (with $p$ and $h_{\#}\rho$ in place of $q$ and $\rho$). For $B\in\mathcal{B}(\mathbb{R}^{p})$, by The One-Particle Marginal of a Probability Measure on the Configuration Space §marginal (in dimension $p$ and in dimension $q$) and Probability Measures on Euclidean Space and Random Vectors: Standing Notation §pushforward,

Claim 4. For $B\in\mathcal{B}(\mathbb{R}^{q})$, by The One-Particle Marginal of a Probability Measure on the Configuration Space §marginal, Claim 1 and Step 0(iv), $(\rho^{\otimes N})^{[1]}(B)=\frac{1}{N}\sum_{k=1}^{N}\rho(B)=N^{-1}(N\rho(B))=\rho(B)$, the last step by Field.

Claim 5. By Particle Blocks: Linearity, Splitting of Inner Products, Product Maps and Diagonal Shifts §inner-product, $\lVert x\rVert^{2}=\sum_{k=1}^{N}\lVert\mathfrak{p}_{k}(x)\rVert^{2}$ for $x\in\mathbb{R}^{qN}$, and each $x\mapsto\lVert\mathfrak{p}_{k}(x)\rVert^{2}$ is Borel and nonnegative by Pairs of Euclidean Points: Coordinate Projections, Pairings, the Product Measure on a Euclidean Space, Borel Norm Functions and Finite Sets §functions and Probability Measures on Euclidean Space and Random Vectors: Standing Notation §borel-maps. By The Second Moment of a Probability Measure on Euclidean Space and the Probability Measures with Finite Second Moment §moment and Step 0(v), $M_{2}(P)=\sum_{k=1}^{N}\int\lVert\mathfrak{p}_{k}\rVert^{2}\,dP$, and by Existence and Uniqueness of Tensor Powers, and the Average of the Block Marginals §average with $f=\lVert\cdot\rVert^{2}$, $M_{2}(P^{[1]})=\frac{1}{N}\sum_{k=1}^{N}\int\lVert\mathfrak{p}_{k}\rVert^{2}\,dP=\frac{1}{N}M_{2}(P)$. By Step 0(vi), $M_{2}(P)=N\,M_{2}(P^{[1]})$; applied to $\rho^{\otimes N}$ together with Claim 4 this gives $M_{2}(\rho^{\otimes N})=N\,M_{2}(\rho)$. By The Second Moment of a Probability Measure on Euclidean Space and the Probability Measures with Finite Second Moment §space and Step 0(vi): if $M_{2}(P)<\infty$ then $M_{2}(P^{[1]})<\infty$, so $P^{[1]}\in\mathcal{P}_{2}(\mathbb{R}^{q})$; and $M_{2}(\rho^{\otimes N})<\infty$ exactly when $M_{2}(\rho)<\infty$.

Claim 6. The map $g^{\oplus}$ is Borel and, by Particle Blocks: Linearity, Splitting of Inner Products, Product Maps and Diagonal Shifts §inner-product in $\mathbb{R}^{pN}$ and Particle Blocks: Linearity, Splitting of Inner Products, Product Maps and Diagonal Shifts §product-map, $\lVert g^{\oplus}(x)\rVert^{2}=\sum_{k=1}^{N}\lVert\hat{\mathfrak{p}}_{k}(g^{\oplus}(x))\rVert^{2}=\sum_{k=1}^{N}\lVert g(\mathfrak{p}_{k}(x))\rVert^{2}$. The map $f=\lVert g(\cdot)\rVert^{2}:\mathbb{R}^{q}\to[0,\infty)$ is Borel by Pairs of Euclidean Points: Coordinate Projections, Pairings, the Product Measure on a Euclidean Space, Borel Norm Functions and Finite Sets §functions and Probability Measures on Euclidean Space and Random Vectors: Standing Notation §borel-maps, and so is each $f\circ\mathfrak{p}_{k}$. By Step 0(v) and Existence and Uniqueness of Tensor Powers, and the Average of the Block Marginals §average,

and Step 0(vi) gives the identity. Now let $Z\in\mathcal{B}(\mathbb{R}^{q})$ with $P^{[1]}(Z)=\frac{1}{N}\sum_{k=1}^{N}P(\mathfrak{p}_{k}^{-1}(Z))=0$. Multiplying by $N$ (Field) gives $\sum_{k=1}^{N}P(\mathfrak{p}_{k}^{-1}(Z))=0$ with nonnegative real summands, so $P(\mathfrak{p}_{k}^{-1}(Z))=0$ for every $k\in[N]$ by claim 5 of Properties of Finite Sums. Let $A_{m}=\mathfrak{p}_{m}^{-1}(Z)$ for $m\in[N]$ and $A_{m}=\varnothing$ for $m\in\mathbb{N}\setminus[N]$; then $\bigcup_{m\in\mathbb{N}}A_{m}=\bigcup_{k=1}^{N}\mathfrak{p}_{k}^{-1}(Z)\in\mathcal{B}(\mathbb{R}^{qN})$ and every $P(A_{m})=0$, so every partial sum of $(P(A_{m}))_{m\in\mathbb{N}}$ is $0$ by Step 0(iv) and the sum of the sequence is $0$ (Measure, Measure Space, and Probability Measure). By claim 4 of Basic Properties of a Measure, $P\bigl(\bigcup_{k=1}^{N}\mathfrak{p}_{k}^{-1}(Z)\bigr)\le0$, hence it is $0$.

Claim 7. The translation $\tau_{a}:\mathbb{R}^{q}\to\mathbb{R}^{q}$, $\tau_{a}(y)=y+a$, is Borel by The Space of Square-Integrable Random Vectors is a Real Hilbert Space; Its Laws Have Finite Second Moment; Constants and Translations §constants, read in dimension $q$ as in Wasserstein Spaces, Random Vectors, Vector Fields and Symmetric Matrices in Every Dimension: Standing Notation §dimensions. For $x\in\mathbb{R}^{qN}$, by Particle Blocks of the Configuration Space: Block Maps, Configurations, Product Maps and Diagonal Points §product-map, Particle Blocks: Linearity, Splitting of Inner Products, Product Maps and Diagonal Shifts §linear (with $t=1$) and Particle Blocks of the Configuration Space: Block Maps, Configurations, Product Maps and Diagonal Points §diagonal,

So $\tau_{a^{\oplus}}=(\tau_{a})^{\oplus}$, and Claim 3 with $q$ in place of $p$ and $h=\tau_{a}$ gives $\bigl((\tau_{a^{\oplus}})_{\#}P\bigr)^{[1]}=(\tau_{a})_{\#}P^{[1]}$ and $(\tau_{a^{\oplus}})_{\#}\rho^{\otimes N}=\bigl((\tau_{a})_{\#}\rho\bigr)^{\otimes N}$.