(related to Proposition: Monotonically Increasing Property of Probability Distributions)
We first show that the probability distribution is monotonically increasing. Assume \(x < y\). Then, we have for the left-open intervals \((-\infty,x]\) and \((-\infty,y]\) that \((-\infty,x]\subset (-\infty,y]\). For the events \(X\le x\) and \(X\le y\), we gain by virtue of the probability of included event the required property \[x < y\Longrightarrow p(X\le x)\le p(Y\le y).\]
Note that following the definition of a random variable, \(X\) is a total map. Therefore, \(X\) will map every outcome of the given random experiment to the set of real numbers \(\mathbb R\). In particular, there is some real number \(x_\emptyset\in\mathbb R\), to which \(X\) maps the impossible realization \(\emptyset\):
\[X(\emptyset)=x_\emptyset.\]
For any real sequence \((x_n)_{n\in\mathbb N}\) tending to minus infinity, \(\lim_{n\to\infty} x_n=-\infty\), there exists an index \(N_\emptyset\in\mathbb N\) such that \[x_n < x_\emptyset\quad\quad\text{for all }n \ge N_\emptyset.\]
It follows from \((i)\) and the probability of the impossible event that
\[p(X\le x_n)\le p(X\le x_\emptyset)=0\quad\quad\text{for all }n \ge N_\emptyset.\]
Because of the axiom telling us that every probability is a non-negative number, we get
\[0\le p(X\le x_n)\le p(X\le x_\emptyset)=0\quad\quad\text{for all }n \ge N_\emptyset,\]
or
\[\lim_{x\to-\infty} p(X\le x)=0.\]
Similarly to \((ii)\), since \(X\) is a total map, there is some real number \(x_\Omega\in\mathbb R\), to which \(X\) maps the certain realization \(\Omega\):
\[X(\Omega)=x_\Omega.\]
For any real sequence \((x_n)_{n\in\mathbb N}\) tending to plus infinity, \(\lim_{n\to\infty} x_n=+\infty\), there exists an index \(N_\Omega\in\mathbb N\) such that \[x_\Omega < x_n\quad\quad\text{for all }n \ge N_\Omega.\]
It follows from \((i)\) and the probability of the certain event that
\[1=p(X\le x_\Omega)\le p(X\le x_n)\quad\quad\text{for all }n \ge N_\Omega.\]
Because of the axiom telling us that the probability is the certain event equals \(1\), we get
\[1=p(X\le x_\Omega)\le p(X\le x_n)=1\quad\quad\text{for all }n \ge N_\Omega.\]
or
\[\lim_{x\to+\infty} p(X\le x)=1.\]
