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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 $mathcal{S}equiv {(a,b,c)in mathbb{R}^3text{ s.t. } a=b+c}$.

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 $$B(b,c)equiv {(x,y,z)text{ s.t. } x> b+c, yleq b, zleq c}$$ and $$Q(b,c)equiv {(x,y,z)text{ s.t. } xleq b+c, y>b, z>c}$$

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)inmathbb{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)inmathbb{Q}^2}B(b,c)={(x,y,z)inmathbb{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 $pin [y,y+epsilon/2)cap mathbb{Q}$ and $qin[z,z+epsilon/2)capmathbb{Q}$. 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)inmathbb{Q}^2}Q(b,c)={(x,y,z)inmathbb{R}^3|x<y+z}=:A_2, .
end{equation}

Together we get that $S^c=A_1cup A_2$. 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.