Proof of The Ordered Time Simplex: Borel Measurability and Volume

Borel measurability. For a real number $c$ and $j\in\{1,\dots,k\}$, the coordinate sets $\{t\in\mathbb{R}^k:t_j<c\}$, $\{t:t_j>c\}$, and $\{t:t_j\le c\}$ are Borel rectangles in the sense of claim 1 of Finite Products of Lebesgue Measure and Coordinate Integration on $\mathbb{R}^l$, all factors being $\mathbb{R}$ except the $j$-th, which is an interval, a Borel set; hence they belong to $\mathcal{B}_k$. For $j<k$, $\{t:t_j<t_{j+1}\}=\bigcup_{n\ge1}\bigcup_{z}\Bigl(\{t:t_j<z/n\}\cap\{t:t_{j+1}>z/n\}\Bigr),$ the inner union taken over all integers $z$: if $t_j<t_{j+1}$, the Archimedean property yields $n\ge1$ with $1/n<t_{j+1}-t_j$, and then the least integer $z$ exceeding $n\,t_j$ satisfies $t_j<z/n\le t_j+1/n<t_{j+1}$. The integers may be listed as the sequence $0,1,-1,2,-2,\dots$, so both unions are countable and the displayed set lies in $\mathcal{B}_k$. Therefore $D_k(T)=\{t:t_1>0\}\cap\{t:t_k\le T\}\cap\bigcap_{j=1}^{k-1}\{t:t_j<t_{j+1}\}\in\mathcal{B}_k.$

Volume. We prove by induction on $k$: for every real $u>0$, $\lambda_k(D_k(u))=u^k/k!$, where $D_k(u)$ denotes the ordered time simplex with horizon $u$. We note first that, in every dimension $k\ge1$, the variant $D_k^\circ(u)$ with last inequality strict ($t_k<u$) has the same measure: $D_k(u)\setminus D_k^\circ(u)$ is contained in the Borel rectangle whose $j$-th factor is $\mathbb{R}$ for $j<k$ and whose $k$-th factor is the interval $\{u\}$, of Lebesgue measure $0$ by claim 4 of Existence of Lebesgue Measure on the Real Line; by the rectangle values of claim 1 of Finite Products of Lebesgue Measure and Coordinate Integration on $\mathbb{R}^l$, whose product conventions give $0$ for any product in $[0,\infty]$ containing the factor $0$, the rectangle has $\lambda_k$-measure $0$.

For $k=1$: $\lambda_1=\lambda$ and $\lambda((0,u])=u$ by claim 4 of Existence of Lebesgue Measure on the Real Line.

For $k\ge2$: under the identification of Finite Products of Lebesgue Measure and Coordinate Integration on $\mathbb{R}^l$, $\lambda_k=\lambda_{k-1}\otimes\lambda$, a product of $\sigma$-finite measures, and the Tonelli theorem applied to the indicator of $D_k(u)$ gives $\lambda_k\bigl(D_k(u)\bigr)=\int_{\mathbb{R}}\lambda_{k-1}\bigl(\{t'\in\mathbb{R}^{k-1}:(t',s)\in D_k(u)\}\bigr)\,d\lambda(s).$ The section is $D_{k-1}^\circ(s)$ for $0<s\le u$ and empty otherwise, so by the inductive hypothesis and the null-boundary remark the integrand equals $\mathbf{1}_{(0,u]}(s)\,s^{k-1}/(k-1)!$, which for $k\ge2$ equals the zero extension of the continuous function $s\mapsto s^{k-1}/(k-1)!$ on $[0,u]$ (both vanish at $s=0$). By claims 2 and 3 of Restricted Lebesgue Measure and Integral Toolkit on a Compact Interval, $\int_{\mathbb{R}}\mathbf{1}_{(0,u]}(s)\,\frac{s^{k-1}}{(k-1)!}\,d\lambda(s)=\int_0^u\frac{s^{k-1}}{(k-1)!}\,ds=\frac{u^k}{k!},$ the Riemann integral evaluated by the fundamental theorem of calculus with base point $0$ and the continuously differentiable antiderivative $s\mapsto s^k/k!$, whose derivative is $s\mapsto s^{k-1}/(k-1)!$ (differentiation of the monomial, by the product rule and induction). This completes the induction; the case $u=T$ is the assertion.