Actual finite Euclidean polynomial measure and moments

Since $U(q)\ge 0$ and $\lambda \ge 0$, [F5] gives $0<\exp (-\lambda U(q))\le 1$ for every $q$, so $Z(\lambda )\le 1$. For each positive integer $n$, let $E_n=\{q:\exp (-\lambda U(q))\ge 1\mathrm{/}n\}$. These are Borel and their union is ${\mathbb{R}}^N$, because the integrand is strictly positive at every $q$. Since ${\gamma }_C$ is a probability measure, not all $E_n$ can have measure zero: otherwise countable subadditivity would give ${\gamma }_C({\mathbb{R}}^N)=0$. Thus some $E_n$ has positive measure, and $Z(\lambda )\ge {\gamma }_C(E_n)\mathrm{/}n>0$.

Fix $r\ge 0$. For $s=0,1,\dots ,r$, [F6] and [F7] give the pointwise derivative $(-1{)}^{s}P(q)U(q{)}^s\exp (-\lambda U(q))$ and its next derivative $(-1{)}^{s+1}P(q)U(q{)}^{s+1}\exp (-\lambda U(q))$. If ${\lambda }_0>0$ and $\mathrm{\mid }h\mathrm{\mid }<{\lambda }_0\mathrm{/}2$, [F8] bounds the difference quotient of the $s$th derivative by $\mathrm{\mid }P(q)\mathrm{\mid }U(q{)}^{s+1}\exp (-{\lambda }_{0}U(q)\mathrm{/}2)$, since every intermediate parameter is at least ${\lambda }_0\mathrm{/}2$ and [F5] makes the exponential increasing. Put $x=U(q)$. If $x=0$ the scalar factor is zero; if $x>0$, put $y={\lambda }_{0}x\mathrm{/}4$. The power-series estimate in [F5] at $2y$ gives $\exp (2y)\ge (2y{)}^{s+1}\mathrm{/}(s+1)!$, so its reciprocal gives $x^{s+1}\exp (-{\lambda }_{0}x\mathrm{/}2)\le (2\mathrm{/}{\lambda }_0{)}^{s+1}(s+1)!\mathrm{.}$ Thus these difference quotients are bounded by a finite constant times $\mathrm{\mid }P(q)\mathrm{\mid }$, integrable by [F2]. [F3] permits differentiation under the integral inductively through order $r$, giving $N_P^{(r)}(\lambda )=(-1{)}^r\int P{U}^r\exp (-\lambda U)d{\gamma }_C$ for $\lambda >0$. Taking $P=1$ gives the formula for $Z$.

The nonnegative Borel density $\exp (-\lambda U)\mathrm{/}Z(\lambda )$ has integral one by the definition of $Z$ and step 1.1. By [F4], it defines a Borel probability measure ${\mu }_{\lambda }$. For any polynomial $P$, [F2] gives $\int \mathrm{\mid }P\mathrm{\mid }d{\gamma }_C<\mathrm{\infty }$; since [F5] gives $\exp (-\lambda U)\le 1$ and step 1.1 gives $Z(\lambda )>0$, $\int \mathrm{\mid }P\mathrm{\mid }d{\mu }_{\lambda }\le Z(\lambda {)}^{-1}\int \mathrm{\mid }P\mathrm{\mid }d{\gamma }_C<\mathrm{\infty }\mathrm{.}$

At $\lambda =0$, use right limits and right difference quotients. For $r=0$, dominated convergence [F3] with dominator $\mathrm{\mid }P\mathrm{\mid }$ and [F5] gives $N_P(\lambda )\to N_P(0)$ as $\lambda \downarrow 0$; taking $P=1$ gives the same for $Z$. Now fix $r\ge 1$. For each $s=0,1,\dots ,r-1$ and $h>0$, [F8] bounds the difference quotient of $(-1{)}^{s}P{U}^s\exp (-\lambda U)$ at zero by $\mathrm{\mid }P\mathrm{\mid }{U}^{s+1}$, since [F5] gives $0<\exp (-cU)\le 1$ for $c\ge 0$. This bound is integrable by [F2]. Dominated convergence [F3] inductively gives $N_{P,+}^{(r)}(0)=(-1{)}^r\int P{U}^{r}d{\gamma }_C$; applying [F3] to the formula in step 1.2 as $\lambda \downarrow 0$ shows these derivatives converge continuously to their endpoint values. Taking $P=1$ gives the same assertions for $Z$.