Finding a probability measure to a $2$-dimensional distribution function

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Suppose $h(x)=0$ for $x<0$ $h(x)=x$ for $0\leq x\leq 1$ and $h(x)=1$ for $x\geq 1$. Now let $F(x,y)=h(x)h(y)$. Explicitly $h(x)=x\cdot 1_{[0,1)}+1_{[1,\infty)}$. Or at least that is what I came up with. Intuitively, $h(x)$ is the distribution function of the uniform distribution on $[0,1]$, but I know,that for higher dimensional distribution, not every distribution function correspondents to a probability measure, at least not without a generalized monotonicity. So how do I conclude, that my $2$-dimensional distribution function is the distribution function of the unfiorm distribution on $[0,1]\times[0,1]$?

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This is a long comment, which is better to put in this section for continuity.

In $\mathbb{R}^2$ (or higher dimensions) , one typically considers right continuous functions $F(x,y)$ with nonnegative increments: that is:

  1. Denote by $\big((a_1,a_2),(b_1,b_2)\big]$ the box $(a_1,b_1]\times(a_2,b_2]$ in $\mathbb{R}^2$ (then left-bottom corner is the point with coordinates $(a_1,a_2)$; the right-upper corner is the point with coordinates $(b_1,b_2)$).
  2. Define $\Delta_1(a, b)F(s):=F(b,s_2)-F(a,s_2)$, and $\Delta_2(a, b)F(s):=F(s_1,b)-F(s_1,a)$.
  3. The increment of $F$ on the box $\big((a_1,a_2),(b_1,b_2)\big]$ is defined as $$\begin{align} m_F((a_1,b_1),(a_2,b_2)])&=\Delta_2(a_2,b_2)\Delta_1(a_1, b_1) F\\ &=\Delta_2(a_2,b_2)(F(b_1,s)-F(a_1,s))\\ &=F(b_1,b_2)-F(b_1,a_2)-F(a_1,b_1)+F(a_1,a_2) \end{align} $$

If $m_F((a_1,b_1),(a_2,b_2)])\geq0$ for all $a_j\leq b_j$, then $m_F$ extends to a measure in $\mathscr{B}(\mathbb{R}^2)$ (The Lebesgue-Stieltjes mesure induced by $F$)

In your case $F(x,y)=\min(x,1)\min(1,y)\mathbb{1}_{(0,\infty)}(x)\mathbb{1}_{(0,\infty)}(y)$. This satisfies the conditions outlined above and gives you the uniform measure in $(0,1]^2$.

Details of all this (in dimension higher than 1) can be found in say, Klenke, A., Probability Theory: A comprehensive course. The arguments for this kind of result are usually based on Caratheodory extension's theorem.