Let $F: \mathbb{R}^d \rightarrow [0,1]$ is a $d$-dimensional distribution function which satisfies the following:
1) $a\leq b,$ $\Delta_{(a,b]}F$ is defined as $F(x_1,x_2,...,x_d)|_{x_1=a_1}^{b_1}...|_{x_d=a_d}^{b_d} \geq 0$.
2) $lim_{y \downarrow x}F(y) = F(x) (y\downarrow x: y$ goes to $x$ in $\Pi_{i=1}^d[x_i,\infty]$)
3) For all $j$, and for all ($X_1, ..., X_{j-1}, X_{j+1}, ..., X_d$), $lim_{x_j \downarrow-\infty}F(x) =0$, and $lim_{x \rightarrow\infty}F(x) =1$ $(x \rightarrow\infty \Leftrightarrow \forall j\leq d, x_j \rightarrow\infty)$.
I should prove that $F(x_1,x_2,...,x_d)$ is non-decreasing in each variable $x_i$'s. If we limit the focus to a two-dimensional function, the statement might be re-written as:
$x\uparrow \Rightarrow F(x,y) \uparrow$ for all $y$ fixed. On the other hand, $y\uparrow \Rightarrow F(x,y) \uparrow$ for all $x$ fixed.
I know that because the function satisfies those properties above, there exists a $d$-dimensional random vector $X$ on some probability space (perhaps ($\mathbb{R}^d, \mathbb{B}(\mathbb{R}^d)$, $P$)...am I right?). I would like to study from which point I should begin to prove (for example, what if I try $d$=2), and I appreciate anyone who may explain the process per step...!
First, convince yourself that in the calculation of $F(x_1,x_2,...,x_d)|_{x_1=a_1}^{b_1}\cdots|_{x_d=a_d}^{b_d} $, the evaluations can be performed in any order of $x_1, \ldots, x_d$.
To show the distribution function is nondecreasing in, say, $x_k$, write down property (1) with limits on $x_k$ evaluated last: We have $$ F(x_1,x_2,...,x_d)|_{x_1=a_1}^{b_1}\cdots|_{x_d=a_d}^{b_d} |_{x_k=a_k}^{b_k}\ge 0 $$ for every $a_1,\ldots, a_d$ and $b_1,\ldots, b_d$ with $a_1\le b_1$, $a_2\le b_2$, etc. (For the purpose of illustration we assume $k\ne1$ and $k\ne2$.) Plugging the limits for $x_1$, this is equivalent to: $$ \left[F(b_1,x_2,...,x_d)-F(a_1,x_2,...,x_d)\right]|_{x_2=a_2}^{b_2}\cdots|_{x_d=a_d}^{b_d} |_{x_k=a_k}^{b_k}\ge 0.\tag{*} $$ Now let $a_1\to-\infty$ in (*). By property (3a), we obtain $$ F(b_1,x_2,...,x_d)|_{x_2=a_2}^{b_2}\cdots|_{x_d=a_d}^{b_d}|_{x_k=a_k}^{b_k}\ge0$$ for every $a_2,\ldots, a_d$ and $b_1,\ldots, b_d$. Continue in this manner, letting each of the $a$'s (except the last) tend to $-\infty$ in turn, and you'll arrive at the assertion $$ F(b_1, b_2,\ldots,b_{k-1},x_k,b_{k+1}\ldots, b_d)|_{x_k=a_k}^{b_k}\ge 0 $$ for every $b_1,\ldots,b_d$ and every $a_k$ such that $a_k\le b_k$. But this is saying $F$ is nondecreasing in the $k$th slot, and we're done!