Consider a 3-dimensional random vector $(X,Y,Z)$. Let $P_{X,Y,Z}$ be the probability distribution of $(X, Y, Z)$. Assume that $$ \int_{\mathcal{S}}dP_{X,Y,Z}=1 $$ where $\mathcal{S}\equiv \{(a,b,c)\in \mathbb{R}^3\text{ s.t. } a=b+c\}$.
My final goal: I'm interested in rewriting the condition $\int_{\mathcal{S}}dP_{X,Y,Z}=1$ as a collection of zero probability measure conditions on boxes in $\mathbb{R}^3$. The idea is that any box in $\mathbb{R}^3$ not intersecting the plane $\mathcal{S}$ should have probability measure equal to zero. Therefore, if we consider enough of these boxes, we should be able to equivalently rewrite $\int_{\mathcal{S}}dP_{X,Y,Z}=1$.
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 $$Q(b,c)\equiv \{(x,y,z)\text{ s.t. } x\leq b+c, y>b, z>c\}$$
I would like your help to see whether the following claim and proof are correct, if not to fix them, if yes to make them more formal.
Claim: $\int_{\mathcal{S}}dP_{X,Y,Z}=1$ if and only if $P_{X,Y,Z}(B(b,c))=0$ and $P_{X,Y,Z}(Q(b,c))=0$ $\forall(b,c)\in \mathbb{R}^2$.
Proof:
Step 1: it is easy to see that if $\int_{\mathcal{S}}dP_{X,Y,Z}=1$ then $P_{X,Y,Z}(B(b,c))=0$ and $P_{X,Y,Z}(Q(b,c))=0$ $\forall(b,c)\in \mathbb{R}^2$.
Step 2: we now show that if $P_{X,Y,Z}(B(b,c))=0$ and $P_{X,Y,Z}(Q(b,c))=0$ $\forall(b,c)\in \mathbb{R}^2$ then $\int_{\mathcal{S}}dP_{X,Y,Z}=1$.
Firstly, notice that if $P_{X,Y,Z}(B(b,c))=0$ and $P_{X,Y,Z}(Q(b,c))=0$ $\forall(b,c)\in \mathbb{R}^2$ then $$ P_{X,Y,Z}(\cup_{b,c} B(b,c))=0 $$ and $$ P_{X,Y,Z}(\cup_{b,c} Q(b,c))=0 $$
Secondly notice that $\cup_{b,c} B(b,c)$ is the open [?] region above the plane $\mathcal{S}$ and that $\cup_{b,c} Q(b,c)$ is the open [?] region below the plane $\mathcal{S}$. Hence $$ \{\cup_{b,c} B(b,c)\} \cup \{\cup_{b,c} Q(b,c)\} $$ is the region that is complement to $\mathcal{S}$ in $\mathbb{R}^3$.
Therefore, $\int_{\mathcal{S}}dP_{X,Y,Z}=1$.
From the comments below: I understand that the step 2 is wrong from the moment in which I take the union over $(b,c)$ because uncountable. Any hint on what to replace? A limit argument for example?
In the following I will assume, that the random vector is $\mathcal{B}(\mathbb{R}^3)$ measurable. Hence we can actually talk about the probability measure of the boxes you defined.
It actually holds a slightly stronger form of your claim: \begin{equation} P_{(X,Y,Z)}(S)=1 \Leftrightarrow P_{(X,Y,Z)}(B(b,c))=P_{(X,Y,Z)}(Q(b,c))=0 \, \forall \,(b,c)\in\mathbb{Q}^2 \end{equation} Proof: "$\Rightarrow$" clear, since $B(b,c)\cap S=\emptyset=Q(b,c)\cap S$.
"$\Leftarrow$" First we show \begin{equation} \bigcup_{(b,c)\in\mathbb{Q}^2}B(b,c)=\{(x,y,z)\in\mathbb{R}^3|x>y+z\}=:A_1 \, . \end{equation} "$\subseteq$" clear
"$\supseteq$" Let $(x,y,z)\in A_1$, so $x>y+z$ and we can define $\epsilon:=x-(y+z)>0$. Since $\mathbb{Q}$ is dense in $\mathbb{R}$ we can find $p\in [y,y+\epsilon/2)\cap \mathbb{Q}$ and $q\in[z,z+\epsilon/2)\cap\mathbb{Q}$. With this we have $x=y+z+\epsilon>p+q$, so $(x,y,z)\in Q(p,q)$.
Similarly it holds that \begin{equation} \bigcup_{(b,c)\in\mathbb{Q}^2}Q(b,c)=\{(x,y,z)\in\mathbb{R}^3|x<y+z\}=:A_2\, . \end{equation}
Together we get that $S^c=A_1\cup A_2$. Since $A_1$ and $A_2$ are disjoint and both a countable union of nullsets we get \begin{equation} P_{(X,Y,Z)}(S^c)=0\, . \end{equation} qed
P.S. I don't have enough reputation to comment.