Rewrite $ \int_{\mathcal{S}}dP_X=1 $ as conditions on boxes in $\mathbb{R}^d$ 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 $r$.
 A: 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.
