Gaussian density integrals for supplied length measures

Every compact box has finite supplied measure, since it is contained in a larger half-open box. A bounded coordinate face has measure zero: enclose it in a half-open slab of arbitrary positive thickness $\eta$, with the other side lengths fixed and positive. Its measure is at most $C\eta$ by the given volume formula and [F2], and so must be zero. In dimension one this says that singletons are null. Consequently compact boxes have their usual volumes, and disjointifying a finite grid of closed cells only removes null faces.

For a nonnegative bounded Borel Riemann integrand $h$ on $Q$, take the infimum and supremum of $h$ on each cell of any finite grid. The corresponding disjoint-cell simple functions bound $h$, and [F2] and step 1.1 make their integrals exactly the lower and upper Darboux sums. By [F3] and [F4], the integral of $h$ is squeezed to its Riemann value. For signed bounded $h$, add a finite constant to make it nonnegative and subtract that constant's finite integral using [F3]. This proves the compact comparison. For nonnegative Borel $H$ on the whole space, $H{\mathbf{1}}_{[-n,n{]}^d}\uparrow H$, so [F5] gives the cutoff assertion. The grids, extrema, and cutoff sequence are finite or explicitly specified; no countable selection is used.

Put $\varphi (t)=(2\pi {)}^{-1\mathrm{/}2}{e}^{-t^2\mathrm{/}2}$. It is continuous and Borel by [F1] and [F7]. On $[-n,n]$, proper substitution $t=\sqrt{2}u$ in [F6] computes its Riemann integral. As $n\to \mathrm{\infty }$, [F6]'s improper Gaussian value makes this tend to one. Step 2.1 identifies each compact integral with its supplied-measure integral, and [F5] gives $\int \varphi d{\lambda }_1=1$.

Since ${\varphi }^{\mathrm{\prime }}=-t\varphi$ by [F7], compact integration by parts [F6] gives ${\int }_{-n}^n{t}^2\varphi (t)dt={\int }_{-n}^n\varphi (t)dt-2n\varphi (n)\mathrm{.}$ All functions and derivatives here are continuous, so the proper integral hypotheses hold by [F4]. Also $n\varphi (n)\to 0$: for $n\ge 1$, $n^2\mathrm{/}2\ge n\mathrm{/}2$, hence $0\le n{e}^{-n^2\mathrm{/}2}\le n{e}^{-n\mathrm{/}2}\to 0$ by [F7]. Apply step 2.1 and monotone convergence [F5] to the nonnegative $t^2\varphi (t)$ cutoffs. Step 3.1 then gives $\int t^2\varphi (t)d{\lambda }_1=1$.

The bound $\mathrm{\mid }t\mathrm{\mid }\le 1+t^2$ and steps 3.1, 4.1 make $\mathrm{\mid }t\mathrm{\mid }\varphi (t)$ integrable. On each symmetric compact interval, the Riemann integral of $t\varphi (t)$ is zero by oddness, using the proper substitution $t=-u$. Step 2.1 transfers these compact integrals to the supplied measure. Dominated convergence [F5], with majorant $\mathrm{\mid }t\mathrm{\mid }\varphi (t)$, yields $\int t\varphi (t)d{\lambda }_1=0$.

Fix $m\in \mathbb{R}$ and $\sigma =\sqrt{v}>0$. On the explicitly specified intervals $[m-\sigma n,m+\sigma n]$, proper substitution $x=m+\sigma t$ gives compact integrals of $g_{m,v}$, $(x-m{)}^2{g}_{m,v}$ and $(x-m)g_{m,v}$ equal to the corresponding integrals of $\varphi$, ${\sigma }^2{t}^2\varphi$ and $\sigma t\varphi$ on $[-n,n]$. Step 2.1 and monotone convergence give mass one and centered second moment ${\sigma }^2=v$. Since $\mathrm{\mid }x-m\mathrm{\mid }\le 1+(x-m{)}^2$, the centered first moment is absolutely integrable, and the same dominated-cutoff argument as step 5.1 gives centered mean zero. Linearity then gives raw mean $m$. This proves both claims for the supplied measures without constructing any new measure or invoking a choice-dependent normal-law supplier.