Stieltjes measure for infinite rectangles In the first chapter of the book "Probability: Theory and Examples" by Rick Durrett, a function $F:\mathbb{R}^D\to\mathbb{R}$ is said to be a Stieltjes measure function if


*

*$F$ is non-decreasing.


That is, if $x_i\leq y_i$ for all $i\in\{1,\dotsc,D\}$, then $F(\mathbf{x})\leq F(\mathbf{y})$.


*

*$F$ is right-continuous.


That is, for every $\mathbf{a}\in\mathbb{R}^D$, given any $\varepsilon>0$ there exists a $\delta>0$ such that for every $\mathbf{x}\in\mathbb{R}^D$ verifying that $0\leq x_i-a_i\leq\delta$ for all $i\in\{1,\dotsc,D\}$, we have that $0\leq F(\mathbf{x})-F(\mathbf{a})<\varepsilon$.


*For every semiopen finite rectangle $A$, this is, $A:=\prod_{i=1}^{D}(a_i,b_i]$ with $-\infty<a_i<b_i<\infty$, we have that
\begin{equation}
\mu(A):=\sum_{v\in V}\sigma(v)F(v)\geq0
\end{equation}
where $V:=\prod_{i=1}^D\{a_i,b_i\}$ and $\sigma(v):=(-1)^{\#a\text{'s in }v}$.



But, how can we define $\mu$ for the non-finite semiopen rectangles of
  $\mathbb{R}^n$?

The book suggests to "use monotonicity" to extend the definition.
I have read another similar definition of this functions in the book by S.G Taylor "Introduction to Measure and Integration" but seems this gap seems to remain.
Suggested definition (Increasing sequence approach)
Let $A$ be an (possibly infinite) semiopen rectangle. That is $A:=\prod_{i=1}^{D}(a_i,b_i]$ with $-\infty\leq a_i<b_i\leq\infty$.
Lets define $\mu(A):=\lim_{k\to\infty}\mu(A_k)$, being $\{A_k\}_{k=1}^{\infty}$ an increasing sequence of finite semiopen rectangles such that $A=\bigcup_{k=1}^{\infty}A_k$. Notice that we can always build a sequence of this kind.
The suggested definition is non-contradictory
Now we have to check that the definition does not depend on the selected sequence. To prove that, we will let $\{A_k\}_{k=1}^{\infty}$ and $\{B_j\}_{j=1}^{\infty}$ be two sequences fulfilling the properties required by our definition.
Main idea: As $\mu$ is non-negative and increasingly monotone for finite semiopen rectangles, the sequences $\{\mu(A_k)\}_{k=1}^{\infty}$ and $\{\mu(B_j)\}_{j=1}^{\infty}$ are non-negative and increasing. So, if given $A_k$ and given any $\varepsilon>0$ we can find a $B_j$ such that $\mu(B_j)\geq\mu(A_k)-\varepsilon$ (and the other way around) both limits will be equal (this is easy to verify).
So given $A_k:=\prod_{i=1}^{D}(\eta_i,\xi_i]$ lets select a $B_j:=\prod_{i=1}^{D}(\alpha_i,\beta_i]$ such that: 


*

*$\beta_i\geq\xi_i$ for all $i\in\{1,\dotsc,D\}$.

*If $a_i=-\infty$, we take $\alpha_i$ such that $\alpha_i<\eta_i$ for all $i\in\{1,\dotsc,D\}$.

*If $a_i>-\infty$, we take $\alpha_i$ such that $(0\leq)\alpha_i-a_i<\delta$ for all $i\in\{1,\dotsc,D\}$. Also, if $\eta_i\not=a_i$, then we take $\alpha_i$ with the additional rectriction of being $\alpha_i<\eta_k$.


Being the $\delta>0$ above, the one associated with the right-continuousity of $F$ in $\mathbf{a}$ and certain $\varepsilon'>0$ to be determined. Notice that we can always choose a $B_j$ fulfilling this conditions.
Calling $V$ to the set of vertices of $A_k$, we define
\begin{equation}
T:=\{v\in V: \text{ if }v_i=\eta_i\text{ and }a_i>-\infty\text{ then }\eta_i\not=a_i\}
\end{equation}
Notice that, if $v\in T$, then there is a set $I\subset\{1,\dotsc,n\}$ such that $v_i=\xi_i$. Now lets consider the semiopen finite rectangle $\widetilde{A_k}:=\prod_{i=1}^{D}(\widetilde{\eta_i},\beta_i]$, where $\widetilde{\eta_i}:=\alpha_i$ if $i\not\in I$ and $\widetilde{\eta_i}:=\eta_i$ otherwise. Notice that $A_k\subset\widetilde{A_k}$.
Let $v$ be a vertex of $\widetilde{A_k}$, then we will denote by $v'$ the vertex of $B_j$ such that if $v_i=\widetilde{\eta_i}$ then $v_i'=\alpha_i$ and if $v_i=\beta_i$ then $v_i'=\beta_i$. 
Finally, we have that
\begin{equation}
\begin{aligned}
\mu(B_j)-\mu(A_k) \geq& \mu(B_j)-\mu(\widetilde{A_k})=\\=&\sum_{v\in T}\sigma(v)(F(v')-F(v))+\sum_{v\not\in T}\sigma(v)(F(v')-F(v))=\\=&\sum_{v\not\in T}\sigma(v)(F(v')-F(v))\geq\\\geq&- (2^D-|T|)\varepsilon'
\end{aligned}
\end{equation}
Taking $\varepsilon'<\frac{\varepsilon}{2^D-|T|}$ we have that $\mu(B_j)\geq\mu(A_k)-\varepsilon$, which is what we wanted.

The proof sounds good to me, but is so overcomplicated. Is there a
  simpler version?

 A: I sense you might have some confusion regarding the author's intention.
To prove the existence of a unique measure based on the Stieltjes measure function, it is necessary to have a three step process:


*

*to define a measure on semialgrebra $\mathcal{S}$ of semiopen rectangles;

*to extend the measure to the algebra $\bar{\mathcal{S}}$ generated by $\mathcal{S}$ with the help of a theorem in the book (by verifying the measure satisfy certain conditions);

*to finally extend the measure to the $\sigma-$algebra generated by $\mathcal{S}$ with the Caratheodory's Extension Theorem (actually bundled into the second step in the book, i.e. with a theorem covering both step 2 & 3).


My guess your reference of

"use monotonicity" to extend the definition

is about 

letting $\mu(A)=\Delta_AF$ for all finite rectangles and then use monotonicity to extend the definition to $\mathcal{S}_d$. 

This is still in the first step of the three step process: $\Delta_AF$ is only defined for finite rectanges, i.e $-\infty < a_i < b_i < +\infty$, while the semialgebra $\mathcal{S}_d$ contains rectangles with infinite semiopen intervals, i.e. $-\infty \le a_i < b_i \le +\infty$. To keep notations simple, let's discuss the case of $d=1$. All you need to extend is to define, for example, $\mu((a_i, +\infty])= \left(\lim_{x\rightarrow \infty}F(x)\right) - F(a_i)$ (and others involving $\pm\infty$). Monotonicity might be involved in verifying we are not violating any definitions of measures, e.g. non-negativity. A full verification might cost some words but the reasoning should be trivial.
So, as you might be able to see now, what the author mean is probably much more trivial than you'd think.
What you described/proved in your question seemed to be something of the 3rd step of the three-step process described earlier. If so, I would think your definition/extension could be problematic. You can read about the details of the proof of the Caratheodory's Extension Theorem to get the right way to extend.
Hope it helps.
