Take $r\in \mathbb{N}$ and let $d\equiv r+\binom{r}{2}$.
Consider a d-dimensional random vector $X\equiv (X_1,...,X_d)$. Let $P_X$ be the probability distribution of $X$. Assume that $$ \int_{\mathcal{S}}dP_X=1 $$ where $$ \begin{aligned} \mathcal{S}\equiv \{(b_1,b_2,..., b_d)\in \mathbb{R}^{d}: \text{ } & b_{r+1}=b_1-b_2, b_{r+2}=b_1-b_3, ...,b_{2r-1}=b_1-b_r, \\ &b_{2r}=b_2-b_3, ..., b_{3r-3}=b_2-b_r,\\ &...,\\ & b_d=b_{r-1}-b_r\} \end{aligned} $$ For example, when $r=2$ ($d=3$) we have the surface $$ \begin{aligned} \mathcal{S}\equiv \{(b_1,b_2,b_3)\in \mathbb{R}^{3}: \text{ } & b_3=b_1-b_2\}=\{(b_1,b_2,b_3)\in \mathbb{R}^{3}: \text{ } & b_1=b_2+b_3\} \end{aligned} $$ When $r=3$ ($d=6$) we have $$ \begin{aligned} \mathcal{S}\equiv \{(b_1,..., b_6)\in \mathbb{R}^{6}: \text{ } & b_4=b_1-b_2, b_5=b_1-b_3, b_6=b_2-b_3\} \end{aligned} $$
My final goal: I'm interested in rewriting the condition $\int_{\mathcal{S}}dP_X=1$ as a collection of zero probability measure conditions on d-dimensional "boxes" in $\mathbb{R}^d$. The idea is that any box in $\mathbb{R}^d$ not intersecting $\mathcal{S}$ should have probability measure equal to zero. Therefore, if we consider enough of these boxes, we should be able to equivalently rewrite $\int_{\mathcal{S}}dP_X=1$.
When $r=2$ ($d=3$), my goal is achieved by the following claim
Claim: For any two real numbers $(b,c)\in \mathbb{R}^2$, define the boxes $$B(b,c)\equiv \{(x,y,z)\text{ s.t. } x> b+c, y\leq b, z\leq c\}$$ and $$Q(b,c)\equiv \{(x,y,z)\text{ s.t. } x\leq b+c, y>b, z>c\}$$
If $P_{X}(B(b,c))=0$ and $P_{X}(Q(b,c))=0$ $\forall(b,c)\in \mathbb{Q}^2$, then $\int_{\mathcal{S}}dP_{X}=1$.
The proof of the claim is provided here
I would like your help to generalise the claim (and possibly the proof) to any $r$. What I find challenging is defining the relevant boxes for any $r>2$. I really can't see how to generalise the box definitions from $r=2$ to any $r$.
I try to post an answer. It is just an attempt, aiming to encourage comments from your side. The attempt mimics for any $r$ the claim and the proof provided here for the case $r=2$ ($d=3$).
Claim
For any $(\bar{b}, \tilde{b})\in \mathbb{R}^2$, consider the $d$-dimensional boxes in $\mathbb{R}^d$ $$ B_{t,p,q}(\bar{b}, \tilde{b})\equiv \{(z_1,...,z_d)\in \mathbb{R}^d \text{: } z_p \leq \bar{b}, \text{ } z_q\leq \tilde{b}, \text{ } z_t>\bar{b}+\tilde{b} \} $$ and $$ Q_{t,p,q}(\bar{b}, \tilde{b})\equiv \{(z_1,...,z_d)\in \mathbb{R}^d \text{: } z_p >\bar{b}, \text{ } z_q> \tilde{b}, \text{ } z_t\leq \bar{b}+\tilde{b} \} $$ $\forall t \in \{1,...,r-1\}$ and $\forall (p,q)\in \{(t+1,r), (t+2, r+1),...,(r, d)\}$. Let $\mathbb{Q}$ denote the set of rational numbers. If \begin{equation} \label{integral} \begin{aligned} P_{X}(B_{t,p,q}(\bar{b}, \tilde{b}))=& P_{X}(Q_{t,p,q}(\bar{b}, \tilde{b}))=0 \\ & \text{$\forall t \in \{1,...,r-1\}$, $\forall (p,q)\in \{(t+1,r), (t+2, r+1),...,(r, d)\}$} \end{aligned} \end{equation} $\forall (b, \tilde{b})\in \mathbb{Q}^{2}$, then $P_{X}(\mathcal{S})=1$.
Proof
Step 1: Using the fact that $\mathbb{Q}$ is dense in $\mathbb{R}$, we can show that
$$ \mathcal{S}^c= \overbrace{\bigcup_{(\bar{b}, \tilde{b})\in \mathbb{Q}^2} \Big\{\bigcup_{\substack{\text{$t\in\{1,...,r_1\}$} \\ \text{$(p,q)\in \{(t+1,r),...,(r,d)\}$}}}\{B_{t,q,p}(\bar{b}, \tilde{b}) \cup Q_{t,q,p}(\bar{b}, \tilde{b})\}\Big\}}^{\equiv A} $$ where $\mathcal{S}^c$ denotes the complement of $\mathcal{S}$.
Step 2: Therefore $$ \mathbb{P}(A)=0 \Leftrightarrow \mathbb{P}(\mathcal{S})=1 $$ from which the conclusion of the claim follows.