A question about probability measure on $(\mathbb{R}^d, \mathcal{R}^d)$. In Rick Durrett's book Probability Theory and Examples, there is a theorem regarding the construction of a probability measure on  $(\mathbb{R}^d, \mathcal{R}^d)$.

Theorem 1.1.11. Suppose $F : \mathbb{R}^d \to [0,1]$ satisfies (i) - (iv) given above. Then there is a unique probability measure $\mu$ on $(\mathbb{R}^d, \mathcal{R}^d)$ so that $\mu(A) = \Delta_A F$ for all finite rectangles.
(i) It is nondecreasing, i.e., if $x \le y$ (meaning $x_i \le y_i$ for all $i$) then $F(x) \le F(y)$
(ii) $F$ is right continuous, i.e., $\lim_{y \downarrow x} F(y) = F(x)$ (here $y \downarrow x$ means each $y_i\downarrow x_i$).
(iii) If $x_n \downarrow -\infty$, i.e., each coordinate does, then $F(x_n) \downarrow 0$. If $x_n \uparrow \infty$, i.e., each coordinate does, then $F(x_n) \uparrow 1$.
(iv) $\Delta_A F\ge 0$, where $\Delta_A F = \sum\limits_{v \in V} \text{sgn}(v)F(v)$ and $\text{sgn}(v) = (-1)^{\text{# of } a_i\text{'s in }v}  $
where $A = (a_1,b_1] \times \cdots \times (a_d, b_d]$ and $V = \{a_1, b_1\} \times \cdots \times \{a_d, b_d\}$
For example, when $A = (a_1, b_1] \times (a_2 ,b_2]$, $\Delta_A F = F(b_1, b_2) - F(a_1, b_2) - F_1(b_1, a_2) + F_1(a_1, a_2)$.

Here, I understand everything in the proof except for why this measure is a probability measure. Intuitively, I get it. However, I think we need to show that $F(b_1, b_2, \cdots, b_d) = \mu((-\infty, b_1] \times (-\infty, b_2] \times \cdots (-\infty, b_d])$ rigorously. Then we can use (iii) to prove $\mu(\mathbb{R}^d) = 1$. At least, in the proof, there is no explicit mention of why this is a probability measure. Any help would be very much appreciated!
 A: I'll suppose you've already proven that $\mu$ exists, is unique, and is a measure. Then it remains to prove that $\mu$ is a probability measure. As you point out, it suffices to show that for any $a$:
$$\mu(x\leq a)=\mu\left(\prod_i(-\infty, a_i]\right)=F(a)$$
To show this, express the set $\{x\leq a\}$ as a disjoint union of finite rectangles (for instance by tiling it with cubes), and use the additivity of $\mu$. For example, in two dimensions, we have:
$$\{x\leq a\}=\bigcup_{n,m}(a_1-(m+1), a_1-m]\times(a_2-(n+1),a_2-n]$$
Due to the precise definition of $\Delta_A F$, this sum can be shown to be telescoping and equal to $F(a)$. Applying additivity, the left hand side above becomes
$$\sum_{n, m} \mu\Big(R_{n,m}\Big)=\sum_{n,m} \Delta_{R_{n,m}} F$$
where $R_{n,m}=(a_1-(m+1), a_1-m]\times(a_2-(n+1),a_2-n]$ is the $n,m$-th "tile". When we write out the definition of $\Delta_A F$, the terms of this infinite sum all cancel out except for the initial term $F(a)$, the top right hand corner. Specifically, this sum expands into a sum of terms, each of the form $\pm F(x)$, where $x$ is some grid point of our infinite tiling. These terms can be grouped into groups of four, each group of four corresponding to a given tile $T$, being the expanded form of $\Delta_T F$. Each tile produces the four terms
$$F(t)-F(u)+F(v)-F(w)$$
where $t,u,v$ and $w$ are the corners of that tile, starting from the top right and going clockwise. Therefore:


*

*The gridpoint $a$, the top right hand corner of the infinite rectangle, appears only once in this entire sum, with a factor of $+1$, as the top right hand corner of the top right hand tile.

*Any gridpoint $x$ which appears in the middle of  the infinite rectangle appears four times, once for each of the four tiles of which it is a corner. It bears a factor of $+1$ for the tiles to its top right and bottom left, and a factor of $-1$ for the other two, therefore all of these terms cancel out and the gridpoint in total makes a contribution of $0$ to the infinite sum.

*Any gridpoint on the right hand side of the infinite rectangle appears twice, once with a factor of $-1$ and once with a factor of $+1$, and therefore also drops out of the sum.

*Analogously for gridpoints on the top of the rectangle.


To show this formally of course would be tedious, especially in $n$ dimensions, but could be done.
