Stability gives finite grand normalization and an exponential particle cutoff

Given: ${\mathrm{A}\mathrm{C}}_{\omega }$, a dimension $d\ge 1$, the model in Stable power-tempered pair-potential model, its stability constant $B$, $a,\beta >0$, and $L\ge 1$.

Stability means $U_N\ge -BN$ for every finite configuration and every $N\ge 1$ (Stable power-tempered pair-potential model).

Measure space and integral defines the nonnegative integral as the supremum of simple minorants. This current batch-11 definition is draft text; its exact interface is inspected in the batch dependency evidence.

The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$) states ${\mathrm{A}\mathrm{C}}_{\omega }$; the standard Lebesgue measure and box suppliers used here state this assumption explicitly.

Lebesgue measurable sets, the family $\mathcal{L}({\mathbb{R}}^n)$, and the restricted set function ${\lambda }_n$ defines ${\lambda }_d$ as Lebesgue measure on the Lebesgue sigma-algebra. Assuming countable choice, every Borel subset of ${\mathbb{R}}^n$ is Lebesgue measurable gives $\mathcal{B}({\mathbb{R}}^d)\subseteq \mathcal{L}({\mathbb{R}}^d)$ under ${\mathrm{A}\mathrm{C}}_{\omega }$, so restricting ${\lambda }_d$ to Borel sets gives a measure. A box in ${\mathbb{R}}^n$ with parameters $a_i\le b_i$ is Lebesgue measurable of measure ${\prod }_{i<n}(b_i-a_i)$, whichever of its faces are included gives ${\lambda }_d([0,L{]}^d)=L^d$ and ${\lambda }_d([-k,k{]}^d)=(2k{)}^d$ for integers $k\ge 1$. For sigma-finite factors, the product measure exists, has the rectangle formula, is sigma-finite, and is unique gives the rectangle formula for finite products of sigma-finite measures.

Lebesgue measure is sigma-finite, and every metrically bounded subset of ${\mathbb{R}}^n$ has finite outer measure gives sigma-finiteness of Lebesgue measure. For its Borel restriction, the Borel boxes $[-k,k{]}^d$ ($k\ge 1$) cover ${\mathbb{R}}^d$ and have finite measure by [F2], so the restricted factors are sigma-finite and their finite product measures in the Statement exist.

The real exponential function and the number $e$ by a power series defines $e^x={\sum }_{N=0}^{\mathrm{\infty }}{x}^N\mathrm{/}N!$, with integer powers as in Integer powers $a^m$; The exponential series converges absolutely for every real argument proves convergence for every real $x$.

The exponential addition formula $\exp (x+y)=\exp (x)\exp (y)$ gives $e^x{e}^y=e^{x+y}$, and The exponential function is strictly increasing gives monotonicity of $e^x$.

The factorial $n!$ and the falling factorial $n^{\underset{‾}{k}}$, defined by recursion in $\mathbb{N}$ defines $0!=1$ and the ordinary factorial recursion; Series in the nonnegative extended real line defines the series sum as the supremum of its nonnegative partial sums, allowing $+\mathrm{\infty }$.

The Borel product of R^m and R^n is the Borel sigma-algebra of R^{m+n} identifies the finite product Borel sigma-algebra, and extended-real measurability means measurability of preimages of extended-real Borel sets (Extended-real-valued measurable functions). Coordinate-difference maps are continuous, hence Borel under ${\mathrm{A}\mathrm{C}}_{\omega }$ (Continuous functions on Euclidean spaces are Borel measurable); composition with the Borel potential preserves measurability (Composition with a Borel measurable outer map preserves measurability). Finite sums of the resulting extended-real measurable terms are measurable whenever defined (Closure properties of measurable functions used by the integral). On finite energy values $u$, $u\mapsto e^{-\beta u}$ is continuous; assigning the value $0$ at $u=+\mathrm{\infty }$ remains Borel because $\{+\mathrm{\infty }\}$ is Borel (The extended real line $\bar{\mathbb{R}}=\mathbb{R}\cup \{-\mathrm{\infty },+\mathrm{\infty }\}$, its order, and the arithmetic that is left undefined, The Borel sigma-algebra on the extended real line, The exponential function is strictly increasing). The box indicator is measurable, so the integrand extended by zero off ${\mathrm{\Lambda }}_L^N$ is product-measurable.

Integer part: for every real $x$ there is exactly one integer $m$ with $m\le x<m+1$ gives, for every real $x$, an integer $\lfloor x\rfloor$ with $\lfloor x\rfloor \le x<\lfloor x\rfloor +1$.