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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!

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    $\begingroup$ Points (i) to (iv) are the hypotheses of the theorem, right? You don't seem to have quoted the actual proof. $\endgroup$
    – Jack M
    Jan 31 '20 at 16:28
  • $\begingroup$ No, I didn't include any of the proof because the proof is all about the existence and uniqueness of the measure $\mu$ satisfying the conditions (i) - (iv). $\endgroup$ Jan 31 '20 at 16:47
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    $\begingroup$ Obviously the intuition here is that $F$ is the cumulative distribution function for $\mu$, so that $F(a)=\mu(x\leq a)$, where $\leq$ is interpreted coordinate-wise. The definition of $\Delta_AF$ is just the formula for the measure of the finite rectangle $A$ given the measures of the infinite rectangles with a vertex at each corner of $A$. So you have the formula for $\mu$ on finite rectangles, and you need to derive the formula for $\mu$ on infinite rectangles. And you already know that $\mu$ is a measure, so we can use that. Is that correct? $\endgroup$
    – Jack M
    Jan 31 '20 at 23:39
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    $\begingroup$ I'm not sure these assumptions are strong enough. If we take say $F(x, y)=\frac1{2\pi}\text{arctan}y+\frac12$, this satisfies all of the assumptions, yet $\Delta_AF=0$ for all $A$, so $F$ is certainly not a cumulative probability function. $\endgroup$
    – Jack M
    Feb 4 '20 at 8:42
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    $\begingroup$ I tracked down the book, and I believe you misquoted it. At least in the first edition, the author has a different point (iii) to yours. In my copy, the assumption is that $F(x)\to1$ as $x\to\infty$, but, $F(x)\to0$ as only one of the $x_i$, not all, goes to $-\infty$. $\endgroup$
    – Jack M
    Feb 4 '20 at 23:42
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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:

  1. 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.
  2. 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.
  3. 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.
  4. 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.

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  • $\begingroup$ Thanks for your answer. Actually, I also thought about this. However, the tricky part for me is $-\infty$ part. For example, in $\mathbb{R}^2$, $\mu(\{x \le a, y \le b\}) = \lim\limits_{m,n \to \infty} \{F(a,b) - F(a-m, b) - F(a, b-n) + F(a-m, b-n)\}$. How can I show $F(a-m, b) \to 0$ as $m \to \infty$ and $F(a, b-n) \to 0$ as $n \to \infty$? Because of $(iii)$, I know $F(a-m, b-n) \to 0$ as $m$, $n \to \infty$. $\endgroup$ Feb 3 '20 at 19:29
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    $\begingroup$ What you're suggesting if different to what I wrote. You're talking about covering an infinite rectangle with a larger and larger rectangle, rather than tiling it with infinitely many rectangles. The approach you're describing seems more difficult to me. $\endgroup$
    – Jack M
    Feb 3 '20 at 20:00
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    $\begingroup$ That equation is what you need to show. The union on the left hand side is equal, as a set, to the set $\{x\leq a\}$ - the big infinite rectangle. The fact that its measure is equal to $F(a)$ ($a$ is the top right corner of the infinite rectangle) is what you need to prove. To do so, what I suggest is to notice that measure is equal to the sum of the measures of the infinitely many "tiles", apply the formula given by your assumptions to calculate those measures, and notice that the sum is telescoping. $\endgroup$
    – Jack M
    Feb 3 '20 at 20:13
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    $\begingroup$ @withgrace1040 I detailed the telescoping argument. $\endgroup$
    – Jack M
    Feb 3 '20 at 21:02
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    $\begingroup$ You're right, there is an additional technical difficulty here. What we need to show is that the value of our infinite sum does not change under permutations of its terms. Showing that this is the case may end up being as difficult or more as the approach that you suggested earlier, so maybe my answer isn't as clever as I thought. $\endgroup$
    – Jack M
    Feb 3 '20 at 22:41

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