Proof of the limit of the cumulative distribution function at infinity

Proof

1. Let $X$ be a real random variable on a probability space $(\Omega, \mathcal{F}, P)$, with cumulative distribution function $F_X(t) = P(X \le t)$.
2. For each integer $n \ge 1$ let $A_n = \set{X \le n}$. It is an event, because $(-\infty, n]$ is a Borel set, and $P(A_n) = F_X(n)$.
3. The events increase: if $X(\omega) \le n$, then $X(\omega) \le n + 1$. So $A_n \subseteq A_{n+1}$.
4. Their union is $\Omega$. Let $\omega \in \Omega$. The value $X(\omega)$ is a real number, so by the Archimedean property there is $n \ge 1$ with $n > X(\omega)$. Then $\omega \in A_n$.
5. By continuity from below, $F_X(n) = P(A_n) \to P(\Omega) = 1$ as $n \to \infty$.
6. Let $\varepsilon > 0$. By the ε-N form of the limit and step 5, there is an integer $N \ge 1$ with $\abs{F_X(N) - 1} < \varepsilon$, that is, $F_X(N) > 1 - \varepsilon$.
7. Let $x \ge N$. The function $F_X$ is nondecreasing, so $F_X(x) \ge F_X(N) > 1 - \varepsilon$. Also $F_X(x) \le 1$, because $F_X(x)$ is a probability.
8. So $\abs{F_X(x) - 1} < \varepsilon$ for all $x \ge N$. Since $\varepsilon > 0$ was arbitrary, $\lim_{x \to +\infty} F_X(x) = 1$.