Proof of Variance of the Empirical Mass of a Borel Set under a Tensor Power

Each result cited is universally quantified over the data in its own statement.

Step 0 (conventions). In real arithmetic the natural number $N$ stands for its image $\iota_{\mathbb{R}}(N)=\sum_{k=1}^{N}1$ under the canonical map; by claim 3 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field it is positive, so $N^{-1}$ exists and $0<N^{-1}$, and $\frac{1}{N}$ means $N^{-1}$. For every real $\lambda$, claim 3 of Properties of Finite Sums gives $\sum_{k=1}^{N}\lambda=\lambda\sum_{k=1}^{N}1=\lambda N$. Write $Q=\rho^{\otimes N}\in\mathcal{P}(\mathbb{R}^{qN})$ (The Tensor Power of a Probability Measure on Euclidean Space §tensor) and $g(x)=\mu^{N}_{x}(B)$. For $k\in[N]$ the block map $\mathfrak{p}_{k}$ is Borel by Particle Blocks: Linearity, Splitting of Inner Products, Product Maps and Diagonal Shifts §linear, so $A_{k}=\mathfrak{p}_{k}^{-1}(B)\in\mathcal{B}(\mathbb{R}^{qN})$, and $\mathbf{1}_{B}(\mathfrak{p}_{k}(x))=\mathbf{1}_{A_{k}}(x)$ for every $x$.

Step 1 ($g$ is Borel with values in $[0,1]$). By Basic Properties of Empirical Measures: Values, Integrals, Push-Forwards, Second Moment, and Lipschitz Dependence on the Configuration §values, $g(x)=N^{-1}\sum_{k=1}^{N}\mathbf{1}_{A_{k}}(x)$ for every $x\in\mathbb{R}^{qN}$. The indicators $\mathbf{1}_{A_{k}}$ are Borel by claim 1 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions, so their linear combination $g$ is Borel by claim 2 of that lemma. Since $\mu^{N}_{x}\in\mathcal{P}(\mathbb{R}^{q})$ by The Empirical Measure of a Configuration of N Particles §empirical, $0\le g(x)\le\mu^{N}_{x}(\mathbb{R}^{q})=1$ by Measure, Measure Space, and Probability Measure and claim 2 of Basic Properties of a Measure. In the same way $c=\rho(B)$ is a real number with $0\le c\le1$.

Step 2 (the pair integrals). For $k\in[N]$, Tensor Powers and One-Particle Marginals: Particle Laws, Product Integrals, Push-Forwards, Moments, Product Maps and Diagonal Shifts §particle-laws and the definition of the push-forward in Probability Measures on Euclidean Space and Random Vectors: Standing Notation §pushforward give $Q(A_{k})=((\mathfrak{p}_{k})_{\#}Q)(B)=\rho(B)=c$. Let $k,l\in[N]$ with $k\ne l$. The pairing $(\mathfrak{p}_{k},\mathfrak{p}_{l}):\mathbb{R}^{qN}\to\mathbb{R}^{q+q}$ is Borel by Pairs of Euclidean Points: Coordinate Projections, Pairings, the Product Measure on a Euclidean Space, Borel Norm Functions and Finite Sets §pairing, and $\mathrm{pr}_{1}\circ(\mathfrak{p}_{k},\mathfrak{p}_{l})=\mathfrak{p}_{k}$, $\mathrm{pr}_{2}\circ(\mathfrak{p}_{k},\mathfrak{p}_{l})=\mathfrak{p}_{l}$ by Pairs of Euclidean Points: Coordinate Projections, Pairings, the Product Measure on a Euclidean Space, Borel Norm Functions and Finite Sets §projections; hence $(\mathfrak{p}_{k},\mathfrak{p}_{l})^{-1}\bigl(\mathrm{pr}_{1}^{-1}(B)\cap\mathrm{pr}_{2}^{-1}(B)\bigr)=A_{k}\cap A_{l}$. By Pairs of Euclidean Points: Coordinate Projections, Pairings, the Product Measure on a Euclidean Space, Borel Norm Functions and Finite Sets §product, the set $\mathrm{pr}_{1}^{-1}(B)\cap\mathrm{pr}_{2}^{-1}(B)=\iota(B\times B)$ is Borel with $(\rho\boxtimes\rho)$-measure $\rho(B)\rho(B)=c^{2}$, and $(\mathfrak{p}_{k},\mathfrak{p}_{l})_{\#}Q=\rho\boxtimes\rho$ by Tensor Powers and One-Particle Marginals: Particle Laws, Product Integrals, Push-Forwards, Moments, Product Maps and Diagonal Shifts §particle-laws; therefore $Q(A_{k}\cap A_{l})=c^{2}$. For $k=l$, $A_{k}\cap A_{l}=A_{k}$ has measure $c$.

For $k\in[N]$ let $f_{k}=\mathbf{1}_{A_{k}}-c$ on $\mathbb{R}^{qN}$. For all $k,l\in[N]$ and $x$, $\mathbf{1}_{A_{k}}(x)\mathbf{1}_{A_{l}}(x)=\mathbf{1}_{A_{k}\cap A_{l}}(x)$ (both sides are $1$ if $x\in A_{k}\cap A_{l}$ and $0$ otherwise, by claim 1 of Zero Products and Elementary Identities in a Field), so the field axioms and claim 2 of Zero Products and Elementary Identities in a Field give

The indicator of a Borel set $C$ is a nonnegative Borel function with $\int\mathbf{1}_{C}\,dQ=Q(C)\le1$ by The Integral of an Indicator Function is the Measure of the Set, hence integrable by Measure Spaces and the Lebesgue Integral: Standing Notation §integral. So Linearity of the Lebesgue Integral over a Finite Sum of Integrable Functions §integrable and Linearity of the Lebesgue Integral over a Finite Sum of Integrable Functions §linear (four summands, coefficients $1,-c,-c,c^{2}$) show that $f_{k}f_{l}$ is integrable with

With the values above and $Q(\mathbb{R}^{qN})=1$: $e_{kk}=c-c^{2}-c^{2}+c^{2}=c(1-c)$, and $e_{kl}=c^{2}-c^{2}-c^{2}+c^{2}=0$ for $k\ne l$.

Step 3 (clause 1). Fix $x$ and put $S(x)=\sum_{k=1}^{N}f_{k}(x)$. By claim 2 of Properties of Finite Sums and Step 0, $S(x)=\sum_{k=1}^{N}\mathbf{1}_{A_{k}}(x)-cN$, so by Step 1 and $N^{-1}N=1$, $N^{-1}S(x)=g(x)-c$. Two applications of claim 3 of Properties of Finite Sums (with $\lambda=S(x)$, then $\lambda=f_{k}(x)$) and commutativity give

For each $k$, Linearity of the Lebesgue Integral over a Finite Sum of Integrable Functions §integrable and Linearity of the Lebesgue Integral over a Finite Sum of Integrable Functions §linear show that $h_{k}$ is integrable with $\int h_{k}\,dQ=\sum_{l=1}^{N}e_{kl}=e_{kk}=c(1-c)$, the middle equality by claim 7 of Properties of Finite Sums since $e_{kl}=0$ for $l\ne k$. Applying the same lemma once more, $(g-c)^{2}$ is integrable and

by Step 0. The integral in clause 1 is that of the nonnegative Borel function $(g-c)^{2}$ in $[0,\infty]$; it equals the real integral just computed because the negative part of a nonnegative function is $0$ and its positive part is the function itself (Integrable Function and the Lebesgue Integral). This proves clause 1.

Step 4 (clause 2). Since $Q(\mathbb{R}^{qN})=1$, $(\mathbb{R}^{qN},\mathcal{B}(\mathbb{R}^{qN}),Q)$ is a probability space. On it let $X=|g-c|$, a random variable by claims 2 and 4 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions, and let $Y$ be the constant $1$, a random variable by claim 1 of that lemma. Pointwise $X^{2}=(g-c)^{2}$, because $|y|$ is $y$ or $-y$ (claim 1 of Properties of the Absolute Value in an Ordered Field) and $(-y)(-y)=yy$ (claim 2 of Zero Products and Elementary Identities in a Field). Hence, with expectations as in Expectation, Variance, and Moments, $\mathbb{E}[X^{2}]=c(1-c)/N$ by Step 3 and $\mathbb{E}[Y^{2}]=Q(\mathbb{R}^{qN})=1$ by The Integral of an Indicator Function is the Measure of the Set; both are finite, so $X$ and $Y$ are square-integrable, with $\lVert X\rVert_{2}=\sqrt{c(1-c)/N}$ and $\lVert Y\rVert_{2}=\sqrt{1}=1$, the latter by the uniqueness in Existence and Uniqueness of the Nonnegative Square Root since $0\le1$ and $1\cdot1=1$. As $XY=X$ is nonnegative and integrable (Square-Integrable Random Variables and the Mean-Square Inner Product), $\delta:=\int_{\mathbb{R}^{qN}}|g-c|\,dQ=\mathbb{E}[XY]$ is a nonnegative real number, equal to $|\delta|$ by the definition of the absolute value. Claim 1 of Cauchy-Schwarz and Triangle Inequalities for the Mean-Square Norm gives $\delta\le\sqrt{c(1-c)/N}$, and claim 2 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field together with Existence and Uniqueness of the Nonnegative Square Root gives $\delta^{2}\le c(1-c)/N$.

Finally, $1-c\le1$ by claim 3 of Elementary Arithmetic in an Ordered Field, since $1-(1-c)=c\ge0$; multiplying by $0\le c$ and then by $0<N^{-1}$, claim 5 of that lemma gives $c(1-c)/N\le c/N$. By transitivity, $\delta^{2}\le c/N$, which is clause 2.