Prerequisites

One should be comfortable with basic measure theory concepts. These include the idea of a sigma field, probability measure, and probability space, $\left(X,{\cal F},{\mathbb{P}}\right).$ Additionally, knowledge of random variables is necessary. These topics are covered in Sigma Fields, Probability Measures, and Random Variables

Summary

A real-valued random variable, $X:\left(\Omega, \mathcal{F}, {\mathbb{P}}\right) \to \left(S, \mathcal{S}\right)$ induces a probability measure on $S.$ For the case of real-valued random variables this measure is called the distribution of $X$ and is usually described by its cumulative distribution function, $F_X(x) := {\mathbb{P}}\left( X \leq x \right).$

Probability Distributions

If $X:\left(\Omega, \mathcal{F}, {\mathbb{P}}\right) \to \left(S, \mathcal{S}\right)$ is an $\mathcal{F}$-measuable r.v., then $X$ induces a probability measure on $\mathcal{S}$. For $A \in \mathcal{S},$ set $\mu\left(A\right):={\mathbb{P}}\left(X \in A\right)={\mathbb{P}}\left(X^{-1}(A)\right)$. It is easy to check that $\mu$ thus defined is a probability measure. For example, observe that for countably many disjoint $A_i$'s, $$\begin{align*} \mu\left(\cup_i A_i\right)={\mathbb{P}}\left(X^{-1}(\cup_i A_i)\r... ..._{i}{\mathbb{P}}\left(X^{-1}(A_i)\right)=\sum_i\mu\left(A_i\right). \end{align*}$$
The other properties of a probability measure can be checked in a similar manner.

In the case that the r.v. $X$ is real-valued, we say that that the induced measure is the distribution of X, and describe this measure by its cumulative distribution function (cdf), $F_X(x):={\mathbb{P}}\left( X \leq x \right)$.

Theorem 1   A cdf $F$ of some probability measure on ${\mathbb{R}}$ has the following properties:

1. $F$ is a nondecreasing function of $x$.
2. $\lim_{x\rightarrow\infty}F(x)=1$ and $\lim_{x\rightarrow -\infty}F(x)=0.$
3. $F$ is right continuous, i.e., $\lim_{y\downarrow x}F(y)=F(x).$
4. If $F(x-) = \lim_{y\uparrow x} F(y)$ then $F(x-)= P(X<x).$
5. $P\left(X= x\right) = F(x) - F(x-).$

Proof.

1. Note that for $x\leq y,$ we have $\{ X \leq x \} \subset \{ X \leq y \}$ so by the monotonicity of probability measures, ${\mathbb{P}}\left( X \leq x \right) \leq {\mathbb{P}}\left(X \leq y \right).$
2. Note that as $x \uparrow \infty,$ we have $\{ X \leq x\} \uparrow \Omega.$ Similarly, as $a \downarrow -\infty,$ $\{ X\leq x\} \downarrow \emptyset.$ By continuity of probability measures, the desired result follows.
3. As $y \downarrow x,$ we have $\{ X \leq y \} \downarrow \{X \leq x\}.$ By continuity of probability measures, the result follows.
4. As $y \uparrow x,$ we have $\{ X \leq y \} \uparrow \{X<x\}.$ By continuity of probability measures, the result follows.
5. We have $\{X \leq x\} = \{X=x\} \cup \{X<x\}.$ Since the events $\{X=x\}$ and $\{X<x\}$ are disjoint, this implies ${\mathbb{P}}\left(X \leq x \right) = {\mathbb{P}}\left(X=x \right) + {\mathbb{P}}\left(X<x \right).$ Rearranging, we have ${\mathbb{P}}\left(X=x \right) = {\mathbb{P}}\left(X \leq x \right) - {\mathbb{P}}\left(X<x \right) .$

$\qedsymbol$

Theorem 2   If $F$ satisfies the first three properties of Theorem , then it is the distribution function of some r.v. and there is a unique probability measure on $({\mathbb{R}},\mathcal{R})$ that has $\mu\left((a,b]\right)=F(b)-F(a)$ for all $a,b \in {\mathbb{R}},\ a \leq b$.

Proof. The proof follows the proof of Durrett's Theorem 1.2. Let $F:{\mathbb{R}}\rightarrow(0,1)$ have properties $1,2,3$ in Theorem . We will construct a r.v. defined on $(\Omega,\mathcal{F},{\mathbb{P}})=((0,1),\mathcal{B}((0,1)),\lambda)$, where $\lambda$ denotes Lebesgue measure, and show that it has distribution function $F$.

Let $$\begin{align*} X(\omega) = \sup\{y: F(y)<\omega \}. \end{align*}$$
If we show that $$\begin{align*} \{\omega : X(\omega) \leq x\} = \{ \omega : \omega \leq F(x) \} \end{align*}$$
then the desired result follows immediately since ${\mathbb{P}}(\omega: \omega \leq F(x)) = F(x)$. (Recall that $\lambda$ is Lebesgue measure on (0,1) and that $F$ is increasing, so that $\lambda\left(\omega: \omega \leq F(x)\right) = \sup \{\omega: \omega \leq F(x)\}$.) To check the set equality above, note that if $\omega \leq F(x)$ then $X(\omega) \leq x$, since $x \notin \{y: F(y)<\omega\}$. On the other hand, if $\omega>F(x)$, then since $F$ is right continuous, there exists an $\epsilon>0$ so that $F(x + \epsilon)<\omega$ and $X(\omega) \geq x + \epsilon>x$. $\qedsymbol$

Having proved the existence of a r.v. $X$ with distribution function $F$, the uniqueness can be checked by Dynkin's $\pi-\lambda$ Theorem. See Appendix A.2 in Durrett (in particular, Theorem 2.2) for further details.

References

Durrett, Probability: Theory and Examples (Third Edition), Section 1.2.