Counterexamples of Cumulative Distribution Function ( multidimensional ) For simplicity, we consider 2-dimensional.
We consider the function $F:\mathbb{R}^2\to\mathbb{R}$.
Let $F$ satisfies:


*

*$0\leq F(x_1,x_2)\leq1$ for $(x_1,x_2)\in\mathbb{R}^2$

*$\left\{\displaystyle\begin{matrix}
\displaystyle\lim_{x_1,x_2\uparrow\infty}F(x_1,x_2)=1, &\\
\displaystyle\lim_{x_1\downarrow-\infty}F(x_1,x_2)=0&\text{for } x_2\in\mathbb{R}, \\
\displaystyle\lim_{x_2\downarrow-\infty}F(x_1,x_2)=0&\text{for } x_1\in\mathbb{R}. \\
 \end{matrix}\right.$

*$\forall(x_1,x_2)\in\mathbb{R}:\ \forall\epsilon>0:\ \exists\delta>0\text{ s.t. }\\
\forall(t_1,t_2)\in\mathbb{R}^2:\ 0\leq t_1-x_1<\delta, 0\leq t_2-x_2<\delta\Rightarrow0\leq F(t_1,t_2)-F(x_1,x_2)<\epsilon$
In addition, if $F$ satisfies: 


*$\forall(x_1,x_2)\in\mathbb{R}^2:\ \forall(h_1,h_2)\in\mathbb{R}_+^2:\\
F(x_1+h_1,x_2+h_2)-F(x_1+h_1,x_2)-F(x_1,x_2+h_2)+F(x_1,x_2)\geq0$


then $F$ is Cumulative Distribution Function.
Now, Let $F^{\prime}$ satisfies 1, 2, and 3. And $F^{\prime}$ is monotonically increasing. Does this $F^{\prime}$ satisfy 4? i.e., Is $F^{\prime}$ Cumlative Distribution Function?
No. We can make counterexamples which $F^{\prime}$ don't satisfy 4. This means 4 is stronger than monotonically increasing property.
But I don't know constitution of $F^{\prime}$. Could you give me some counterexamples?

The function
$$
F(x, y):=\begin{cases}
1& \text{if $x+y\ge0$}\\
0& \text{otherwise}
\end{cases}
$$
satisfies conditions (1) and (2). It also meets condition (3), since for any $(x, y)$ there is $h>0$ such that $F(x+h, y+h)=F(x, y)$. However, condition (4) is not satisfied since
$$
F(1,1) + F(0,0) = 1 + 0 < 1 + 1 = F(1,0) + F(0,1).
$$
