Is $H$-measure actually monotonic (at least on hyperrectangles)? I am currently reading Introduction to Copulas by R. B. Nelson. First chapter introduces some theory of 2-monotone functions and I am trying to extend it for $n$-dimensional hyperrectangles as an exercise. It still unclear for me if $H$-measure is a real measure or at least monotone (the property I crave the most). The theory is following:
Let $S_i$ be a family of $n$ subsets of $\overline{\mathbb{R}} = \mathbb{R} \cup \{-\infty, \infty\} $.
Then, we call a function of type $H: \prod^n_{i=1}S_i \to \mathbb{R}$  $n$-place real-valued function.
For every $n$-dimensional hyperrectangle $R$ of form $R = \prod^n_{i=1} [a_i,b_i]$ with $a,b \in \prod^n_{i=1}S_i$ we define its $H$-volume by $$V_H(R) = \Delta^{b_n}_{a_n} \ldots \Delta^{b_2}_{a_2}\Delta^{b_1}_{a_1}H = \sum_{I \in \{ 1,2\}^n} (-1)^{\sum^n_{i = 1} I_i}H[I] $$
where $H[I] = H(x_{I_1},x_{I_2}, \ldots, x_{I_n})$ having $x_{I_k} = a_k$ if $I_k = 1$ and $b_k$ otherwise.
Here is an example for $\mathbb{R}^2$:
$$V_H(R) = H(b_1,b_2) - H(a_1,b_2) - H(b_1,a_2) + H(a_1,a_2)$$
If $\forall R . V_H(R) \ge 0$ then $H$ is called $n$-increasing or quasi-monotone and $V_H(R)$ is called an $H$-measure of $R$. This invokes the idea that $V_H$ is indeed acts as measure, hence it is monotone i. e. $$\forall R,T. R \subset T \Rightarrow V_H(R) \le V_H(T)$$
Now, I think that it's possible to look at this measure as normall integral.
$$V_H(R) = \int_R \mathrm{d}^n\, H(p)$$ And Equality can be shown by multiple application of Stokes' theorem to Lebesgue–Stieltjes integral. To get monotonicity we just need to ensure  $H$ is nondeacrising in every coordinate, which we have in case $H$ is grounded i.e there are some least element $z_i$ in each $S_i$ and $f(\ldots, z_i, \ldots) =0.$  I think we can consider it to be grounded with only little lose of generality.
 A: Thinking about $V_H(R)$ as a result of integration is wrong as $H$ doesn't have to be a proper destribution or defined in any point of $R$ except its vertices. However, it is possible to prove the conjecture only in terms of volumes  and vertices of rectangles. In case of $\mathbb{R}^2$ it is possible to draw the following picture, where $R$ is depicted as white rectangle and $T$ is depicted as grey one and $R \subset T$:

In this case we can write a sum for $V_H(T) - V_H(R)$ . Then we can add and subtract $H(p_k)$ for all the pink points on the picture, This enables us to convert this sum to  volumes of four smaller grey rectangles with some parts of the walls dashed and by definition this volumes must be greater or equel to zero. Hence $V_H(T) - V_H(R) \ge 0$ and $V_H(R)$ is monotonic. This proof can be easily extended in higher dimensions, however explicit enumeration of all extra vertices is too tedious, so we will omit this part.
If  something is wrong with this proof, I would be thankful for your corrections.
