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?