# Criteria for a distribution function to uniquely determine a measure on $(\mathbb{R}^2,\mathcal{B}(\mathbb{R}^2))$

I know that for a function $$F:\mathbb{R}\to[0,1]$$ with the below criteria, $$F$$ is a distribution function for a probability measure $$\mu_F$$. Indeed, we can define $$\mu_F=F(b)-F(a)$$ for all $$a,b\in\mathbb{R},a the Lebesgue-Stieltje measure.

Criteria: $$F$$ has to be monotone increasing, right-continuous and $$F(-\infty)=0$$, $$F(\infty)=1$$.

But when looking at $$(\mathbb{R}^2,\mathcal{B}(\mathbb{R}^2))$$, are there similar criteria to create a measure from a function $$F:\mathbb{R}^2\to[0,1]$$ ? That is, $$F$$ is the distribution function of a probability measure $$\mu$$ on $$(\mathbb{R}^2,\mathcal{B}(\mathbb{R}^2))$$ ?

My thoughts so far:

If $$Z=(X,Y)$$ is a random variable with distribution $$\mu$$, then we want $$F(z)=F(x,y)=P_{\mu}(X\leqslant x,Y\leqslant y)$$ for all $$x,y\in\mathbb{R}$$. This would mean that $$F(-z)\to0\text{ and }F(z)\to1\text{ when }\vert z\vert\to\infty$$ that is $$F(-\infty,-\infty)=0\text{ and } F(+\infty,+\infty)=1$$ This is our first hypothesis. We have to keep the monotone increasing hypothesis, thus I suggest $$F(x_1,y)\leqslant F(x_2,y)\text{ for }x_1\leqslant x_2$$ and $$F(x,y_1)\leqslant F(x,y_2)\text{ for }y_1\leqslant y_2$$ but I am not sure. For $$F$$ being right-continuous, we can relax it to be continuous, or maybe it we can only rely on $$F$$ being continuous from the right and from the top (same kind of idea in 1D but here in 2D).

Now if $$z_1=(x_1,y_1)$$ and $$z_2=(x_2,y_2)$$ are such that $$x_1< x_2$$ and $$y_1< y_2$$, then the rectangle $$(x_1,y_1), (x_2,y_1), (x_2,y_2), (x_1,y_2)$$ has a positive volume, thus the function $$F$$ has to increase in that domain. This would mean that another hypothesis is $$F((x_2,y_2))-F((x_2,y_1))-F((x_1,y_2))+F((x_1,y_1))\geqslant 0$$ for all $$x_1\leqslant x_2, y_1\leqslant y_2$$.

Thus I suggest the following:

There exists a measure $$\mu$$ such that $$F$$ is its distribution function if and only if:

(1) $$F$$ is right-continuous and top-continuous (right-continous on each of its variable)

(2) $$F$$ is increasing: $$F(x_1,y)\leqslant F(x_2,y)\text{ for }x_1\leqslant x_2$$ and $$F(x,y_1)\leqslant F(x,y_2)\text{ for }y_1\leqslant y_2$$

(3) $$F$$ gives a positive value for a rectangle: $$F((x_2,y_2))-F((x_2,y_1))-F((x_1,y_2))+F((x_1,y_1))\geqslant 0$$ for all $$x_1\leqslant x_2, y_1\leqslant y_2$$.

In that case, $$\mu$$ is defined to be $$\mu([x_1,y_1]\times[x_2,y_2])=F((x_2,y_2))-F((x_2,y_1))-F((x_1,y_2))+F((x_1,y_1))$$ on all rectangles of $$\mathbb{R}^2$$, and the sets of all rectangles is a generator of $$\mathcal{B}(\mathbb{R}^2)$$ so we may extend the value of $$\mu$$ for all Borel sets of $$\mathbb{R}^2$$.

I'll try to prove this and see if this works, still any help is appreciated ! Thanks !

The general result is the following, which is an extension of your proposal:

$$F:\mathbb{R}^n\to\mathbb{R}$$ is the distribution function of a random vector $$\mathbf{X}$$ if and only if

• $$F$$ is a Stieltjes measure function, that is:

1. Componentwise non-decreasing.

2. Right continuous.

3. Given a rectangle $$A:=\prod_{i=1}^n(a_i,b_i]$$ where $$-\infty, then $$\mu_F(A):=\sum_{\mathbf{v}\in V}\sigma(\mathbf{v})F(\mathbf{v})\geq 0$$, where $$V:=\prod_{i=1}^n\{a_i,b_i\}$$ and $$\sigma(\mathbf{v}):=(-1)^{\#a's\in \mathbf{v}}$$.

• $$\lim_{\mathbf{x}\to\infty}F(\mathbf{x})=1$$.
• $$\lim_{x_i\to-\infty}F(\mathbf{x})=0$$ for all $$1\leq i\leq n$$.

# A path to the proof

The proof is long and painful, but just as long and painful as the proof for the case $$n=2$$.

1. Prove that the family of sets of the form $$\prod_{i=1}^n(a_i,b_i]$$ where $$-\infty\leq a_i\leq b_i\leq \infty$$ is a semialgebra, that is, the family is closed under finite intersections, and, the complement of a set of the family is a disjoint union of sets in the family. (Easy).
2. Prove that if $$A=\prod_{i=1}^n(a_i,b_i]=\bigoplus_{j=1}^m\prod_{i=1}^n(c_i^j,d_i^j]$$ being $$\bigoplus$$ a disjoint union and $$-\infty, then $$\mu_F(A)=\sum_{j=1}^m\mu_F(c_i^j,d_i^j]$$. (Very Hard!!).

Notation will be extremely important proving this. Also, you may want to prove first that, if $$A$$ is a set of a semialgebra which is a finite union of sets of the semialgebra, then $$A$$ has a regular subdivision formed by sets of the semialgebra. (Which is also very hard!)

1. Prove that if $$A\subset B$$, being $$A=\prod_{i=1}^n(a_i,b_i]$$ and $$B=\prod_{i=1}^n(c_i,d_i]$$ with $$-\infty and $$-\infty, then $$\mu_F(A)\leq\mu_F(B)$$. (Easy).
2. Extend the function $$\mu_F$$ to sets of the form $$A:=\prod_{i=1}^n(a_i,b_i]$$ with $$-\infty\leq a_i\leq b_i\leq \infty$$. One option is to define $$\mu_F(A)=\lim_{k\to\infty}\mu_F(B_k)$$ where $$\bigcup_{k=1}^\infty B_k=A$$, being $$B_k=\prod_{i=1}^\infty (c_i,d_i]$$ with $$-\infty and $$B_k\subset B_{k+1}$$. Prove that the definition is unambiguous. (hard!).
3. Prove that if $$A=\prod_{i=1}^n(a_i,b_i]=\bigoplus_{j=1}^m\prod_{i=1}^n(c_i^j,d_i^j]$$ being $$\bigoplus$$ a disjoint union and $$-\infty\leq a_i\leq b_i\leq\infty$$, then $$\mu_F(A)=\sum_{j=1}^m\mu_F(c_i^j,d_i^j]$$. (Easy).
4. Prove that if $$A=\prod_{i=1}^n(a_i,b_i]=\bigoplus_{j=1}^\infty\prod_{i=1}^n(c_i^j,d_i^j]$$ being $$\bigoplus$$ a disjoint union and $$-\infty\leq a_i\leq b_i\leq\infty$$, then $$\mu_F(A)\leq\sum_{j=1}^\infty\mu_F(c_i^j,d_i^j]$$. (Hard).
5. Prove the Carathéodory's extension theorem (Hard).
6. Conclude that $$\mu_F$$ is a measure over the borelians of $$\mathbb{R}^n$$, and, in particular, a probability measure. (Easy).
7. Prove that $$F$$ is the distribution function, that is, $$F(\mathbf{x})=P(\mathbf{X}^{-1}(\prod_{i=1}^n(-\infty,x_i]))$$, of the random vector $$\mathbf{X}:(\mathbb{R}^n,\beta,\mu_F)\to\mathbb{R}^n$$ such that $$\mathbf{X}=\text{id}$$. (Easy).

The converse is easy, you will have to use the inclusion-exclusion principle, look here for an idea.

You can find some of the proofs in Rick Durrett's book, but those marked with bold are quite difficult and aren't anywhere (or at least I'm not smart enough to find them). If you want some clarifications, feel free to ask for them!

# The hard and "missing" parts (to be continued)

## The extension of $$\mu_F$$ is well defined

Let $$A:=\prod_{i=1}^n(a_i,b_i]$$ be, with $$-\infty\leq a_i\leq b_i\leq \infty$$. Now, lets consider two increasing families of sets $$\{B_n\}_{n=1}^\infty$$ and $$\{C_n\}_{n=1}^\infty$$ such that $$\bigcup_{n=1}^\infty B_n = \bigcup_{n=1}^\infty C_n=A$$, being $$B_n:=\prod_{i=1}^n(c_i,d_i]$$ with $$-\infty and $$C_n:=\prod_{i=1}^n(p_i,q_i]$$ with $$-\infty.

Lets prove that $$\lim_{n\to\infty}\mu_F(B_n)=\lim_{n\to\infty}\mu_F(C_n)$$. As proved before, $$\{\mu_F(B_n)\}_{n=1}^\infty$$ and $$\{\mu_F(C_n)\}_{n=1}^\infty$$ are increasing sequences, so, if for every $$\varepsilon>0$$ and every $$B_n$$ there is a $$C_m$$ such that $$\mu_F(C_m)\geq \mu_F(B_n)-\varepsilon$$, then $$\lim_{n\to\infty}\mu_F(B_n)\leq\lim_{n\to\infty}\mu_F(C_n)$$. This is fundamental for the rest of the proof.

Given $$\varepsilon>0$$ and $$B_n:=\prod_{i=1}^n(c_i,d_i]$$, as $$F$$ is right continuous, for every $$\varepsilon'>0$$ and every vertex $$\mathbf{v}\in B_n$$ there is a $$\delta_{\mathbf{v}}$$ such that is $$0\leq v_i-w_i\leq \delta_{\mathbf{v}}$$ then $$F(\mathbf{w})-F(\mathbf{v})<\varepsilon'$$. Lets define $$\delta:=\min{\delta_{\mathbf{v}}}$$.

We can always find a $$C_m:=\prod_{i=1}^n(p_i,q_i]$$ such that

1. $$q_i\geq d_i$$.
2. If $$a_i\not=c_i$$ then $$p_i.
3. If $$a_i=c_i$$ then $$|p_i-a_i|<\delta$$.

Now, we have that $$$$\mu_F(C_m)-\mu_F(B_n)\geq \mu_F(C_m\cap B_n)-\mu_F(B_n)=\sum_{\mathbf{v}\in V}\sigma(\mathbf{v})(F(\mathbf{w})-F(\mathbf{v}))$$$$ where $$\mathbf{w}$$ if the vertex of $$C_m\cap B_n$$ corresponding to $$\mathbf{v}$$. It is clear that $$w_i\geq v_i$$, so, as $$F$$ is increasing $$$$\sum_{\mathbf{v}\in V}\sigma(\mathbf{v})(F(\mathbf{w})-F(\mathbf{v}))\geq\sum_{\mathbf{v}\in V}F(\mathbf{v})-F(\mathbf{w})$$$$ then, using the right-continuity of $$F$$ we have that $$F(\mathbf{v})-F(\mathbf{w})>-\varepsilon'$$. There are $$2^n$$ vertices, so $$\mu_F(C_m)\geq \mu_F(B_n)-2^n\varepsilon'$$. Taking $$\varepsilon'<\frac{\varepsilon}{2^n}$$ we are done.

• When you say $F$ is right-continuous, does this mean you converge from above for every component ? – Flewer47 Jun 8 '20 at 10:57
• @Flewer47 yes, I mean that for every $\mathbf{x}\in\mathbb{R}^n$ and every $\varepsilon>0$ there is a $\delta>0$ such that if $0<y_i-x_i<\delta$ for every $1\leq i\leq n$ then $|f(\mathbf{y})-f(\mathbf{x})|<\varepsilon$. – Álvaro G. Tenorio Jun 8 '20 at 11:11