Limits of cumulative distribution function I want to know if there is a formal proof of the statement that, given a cumulative distribution function $F_X(x)=P(X\leq x)$, the limits $\lim\limits_{x\to\infty} F_X(x) = 1$ and $\lim\limits_{x\to -\infty} F_X(x) = 0$ hold.
I have tried using the facts that, in every probability space, $P(\emptyset)=0$ and $P(\Omega)=1$. But I haven't figured out a proof yet.
Thanks to anyone.
 A: Let $(\Omega, \mathcal{F}, P)$ be a probability space. Let $X \colon \Omega \to \mathbb{R}$ be a random variable. Also, let $A_x = \{ \omega \in \Omega \colon X(\omega) \leq x \}$.
(In words, $A_x$ is just the set of outcomes such that our random variable $X$ takes a value of at most $x$. Also note that $A_x \in \mathcal{F}$ by the definition of a random variable.)
The distribution function of $X$ is the function $F_X \colon \mathbb{R} \to [0, 1]$ where $F_X(x) = P(X \leq x) = P(A_x)$.
Now that we've made our definitions, let's get down to business.
Claim: $\lim_{x \to \infty} F_X(x) = 1.$
Proof: Let $(x_n)$ be a monotonic sequence of real numbers converging to $\infty$. This means that $A_{x_n} \subseteq A_{x_{n + 1}}$ for all $n$. (Informally, the set of outcomes $\omega \in \Omega$ such that the random variable $X$ takes on a value of no more than $x_n$ can't be any larger than the set of outcomes $\omega$ such that $X$ takes on a value no more than $x_{n + 1}$, since $x_n \leq x_{n + 1}$ for all $n$.)
We therefore have a sequence of sets $A_{x_1} \subseteq A_{x_2} \subseteq A_{x_3} \subseteq \dots$, and it's important to note that $\cup_{n = 1}^{\infty} A_{x_n} = \Omega$. (Why? Because the infinite union is all the $\omega \in \Omega$ such that $X(\omega) < \infty$, which is all of $\Omega$.)
I now need to call on the result that given a sequence of sets $(A_n)$ where $A_1 \subseteq A_2 \subseteq A_3 \subset \dots$, then $\lim_{n \to \infty} P(A_n) = P(\cup_{i = 1}^{\infty} A_i).$
Therefore, 
\begin{align*}
\lim_{x \to \infty} F_X(x) &= \lim_{n \to \infty} F_X(x_n)\\
&= \lim_{n \to \infty} P(A_{x_n})\\
&= P \big( \bigcup_{n = 1}^{\infty} A_{x_n} \big)\\
&= P(\Omega)\\
&= 1.\square
\end{align*}
The proof that $\lim_{x \to -\infty}F_X(x) = 0$ is pretty similar.
