# Proof

### $$(i)$$ Proof of monotonically increasing property

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 $x < y\Longrightarrow p(X\le x)\le p(Y\le y).$

### $$(ii)$$ Proof of $$\lim_{x\to-\infty} p(X\le x)=0$$

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

$\lim_{x\to-\infty} p(X\le x)=0.$

### $$(iii)$$ Proof of $$\lim_{x\to+\infty} p(X\le x)=1$$

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

$\lim_{x\to+\infty} p(X\le x)=1.$

Github: ### References

#### Bibliography

1. Forster Otto: "Analysis 1, Differential- und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983
2. Hedderich, J.;Sachs, L.: "Angewandte Statistik", Springer Gabler, 2012, Vol .14