"[Measure Theory and Probability Theory](https://link.springer.com/book/10.1007/978-0-387-35434-7)" by Athreya and Lahiri introduces Lebesgue–Stieltjes measure construction on $\mathbb{R}^n$ in general in the following way:


Let $f:\mathbb{R}^2\rightarrow \mathbb{R}$ satisfy the following properties:


1. Let $(x\_1,x\_2)\leq (y\_1,y\_2)$ iff $x\_1\leq y\_1$ and $x\_2\leq y\_2$.
If $x\leq y$ in this partial order, then $f(y\_1,y\_2)-f(x\_1,y\_2)-f(y\_1,x\_2)+f(x\_1,x\_2)\geq 0$.
2. $f$ is right-continuous for each coordinate.
3. $x\leq y\ (\forall x\_n\leq y\_n)\implies f(x)\leq f(y)$ (this condition is omitted in the book, I think this might be needed).


Then $\mu((x\_1,x\_2]\times (y\_1,y\_2])=f(y\_1,y\_2)-f(x\_1,y\_2)-f(y\_1,x\_2)+f(x\_1,x\_2)$ defines a locally finite Borel measure on $\mathbb{R}^2$. Conversely, any locally finite Borel measure is generated by such a function.


I have some questions about this construction:


1. I want to see a complete proof of this construction in $\mathbb{R}^n$; the first condition can be defined analogously in $\mathbb{R}^n$. Even in the book, there was no complete proof of this general construction. I am also not sure whether the third condition can be omitted or not.
2. I wonder if there is a one-to-one correspondence between such functions and locally finite Borel measures on $\mathbb{R}^n$.
In the one-dimensional case, a locally finite Borel measure corresponds to a right-continuous increasing function up to a constant. It seems like such a correspondence result, if it exists, should be more restrictive; $f(x,y)=x+y$ yields the zero measure, but any $f(x,y)=F(x)+G(y)$ does so as well.


Any reference or complete construction would be much appreciated.


(The question was originally asked in [stackexchange](https://math.stackexchange.com/questions/4009789/lebesgue-stieltjes-measure-construction-on-mathbbrn).)


[Edited: Added the authors.]



Lebesgue–Stieltjes measure construction on $\mathbb{R}^n$
$\newcommand\R{\mathbb R}$Concerning your question 1: A complete proof of this result (based on a discrete approximation) can be found, for instance, in [Kallenberg's book "Foundations of Modern Probability"](https://www.springer.com/gp/book/9780387953137), where it is given as Corollary 3.26. Another kind of arguments, using e.g. semirings, can be found e.g. in Example 1.54 in [Probability Theory, A Comprehensive Course, by Klenke](https://www.springer.com/gp/book/9781447153610), where the Lebesgue measure over $\R^d$ is constructed; the construction of the Lebesgue–Stieltjes measure is quite similar.


Concerning your question 2: Clearly, if $\mu\_f$ denotes the Lebesgue–Stieltjes measure for a right-continuous function $f\colon\R^d\to\R$ with nonnegative increments (in the natural sense, as defined e.g. in Kallenberg's book), and if $g\colon\R^d\to\R$ is such that
\begin{equation}
\label{star}
\tag{$\*$}
g(x\_1,\dotsc,x\_d)=f(x\_1,\dotsc,x\_d)+\sum\_{j=1}^d h\_j(x\_1,\dotsc,x\_{j-1},x\_{j+1},\dotsc,x\_d)
\end{equation}
for some functions $h\_j\colon\R^{d-1}\to\R$ and all $(x\_1,\dotsc,x\_d)\in\R^d$, then $\mu\_g=\mu\_f$.
Vice versa, if $\mu\_g=\mu\_f$, then the (appropriate $d$-fold) increments of $g-f$ are identically zero, which implies \eqref{star}.


