Suppose that
\((\Omega,\mathcal{F})\) and \((\Omega',\mathcal{F}')\) are measurable spaces,
\(T:\Omega\rightarrow \Omega'\) is measurable-\(\mathcal{F}/\mathcal{F}'\), and
\(\mu\) is a measure on \(\Omega\).
Define a set function \(\mu_T\) (or \(\mu T^{-1}\) or \(\mu \circ T^{-1}\)) on \(\mathcal{F}'\) by \[ \mu_T(A')=\mu(T^{-1}(A')), \mbox{ }A'\in \mathcal{F}'. \hspace{5mm}(T^{-1}(A') \in \mathcal{F}) \]
\(\mu_T\) is nonnegative:
\(\mu_T(\phi)=\mu(\phi)=0\);
\(\mu_T\) is countably additive: if \(A_1', A_2',\ldots\in \mathcal{F}\) are disjoint, then \(T^{-1}(A_1'), T^{-1}(A_2'),\ldots\in \mathcal{F}\) are disjoint.
\(\left(\mbox{간단증명: }\omega\in T^{-1}(A_1')\implies T(\omega)\in A_1'\implies T(\omega)\notin A_2'\implies \omega\notin T^{-1}(A_2')\right)\)
Since \(\mu\) is countably additive, \[ \mu_T(\uplus_n A_n')=\mu(T^{-1}(\uplus_n A_n'))=\mu(\uplus_n T^{-1}(A_n'))=\sum_n \mu( T^{-1}(A_n'))=\sum_n\mu_T(A_n'). \]
Thus, \(\mu_T\) is a measure on \((\Omega', \mathcal{F}')\), and is called the measure induced by \(T\). If \(\mu\) is a finite measure, then so is \(\mu_T\). In particular, if \(\mu\) is a probability measure, then so is \(\mu_T\).
If \((\Omega, \mathcal{F}, P)\) is a probability space, and \(X:\Omega\rightarrow \mathbb{R}\) is a random variable, then the probability measure \(P_X\) induced on \((\mathbb{R},\mathcal{R})\) by \(X\) is called the distribution of \(X\), and in this case we may write any of the following: \[ P_X(H)=P(X^{-1}(H))= P(\{\omega\in\Omega: X(\omega)\in H\})= P(X\in H) \hspace{3mm} \mbox{ for all } H\in \mathcal{R}. \]
The distribution function of a random variable \(X\) is the (cumulative) probability distribution function \(F=F_X\) corresponding to the measure \(P_X\), i.e., \[ F(x)=P_X\left((-\infty,x]\right)=P(X\le x), \hspace{3mm} x\in \mathbb{R}. \]
If \(F\) is a probability distribution function on \(\mathbb{R}\), then the quantile function \(F^{-1}\) is finite, nondecreasing, and left-continuous on (0,1) and satisfies
\(F(x)\ge u\) if and only if \(x\ge F^{-1}(u)\)
\(\iff F(x)<u\) if and only if \(x<F^{-1}(u)\).
\(F^{-1}(0+)=-\infty\) if and only if \(F(x)>0\) for all \(x\in \mathbb{R}\).
\(F^{-1}(1-)=\infty\) if and only if \(F(x)<0\) for all \(x\in \mathbb{R}\).
If \(F\) is a probability distribution on \(\mathbb{R}\), then there exists a probability space \((\Omega, \mathcal{F}, P)\) and a random variable \(X\) defined on \(\Omega\) having distribution function \(F\).
First construction:
Let \(\mu\) be the unique measure on \((\mathbb{R},\mathcal{R})\) satisfying \(\mu((a,b])=F(b)-F(a)\)
(In the given probability space, there exist a unique measure w.r.t the probability distribution function \(F\) by correspondence theorem).
Then \[ \mu((-\infty,x])=\lim_{n\rightarrow \infty} \mu((-n,x])=F(x)-\lim_{n\rightarrow \infty}F(-n)=F(x)-0=F(x), \] and \[ \mu(\mathbb{R})=\lim_{n\rightarrow \infty} \mu((-\infty,n])=\lim_{n\rightarrow \infty}F(n)=1. \]
Thus, \(\mu\) is a probability measure on \((\mathbb{R},\mathcal{R})\).
Now let \((\Omega, \mathcal{F},P)=(\mathbb{R},\mathcal{R},\mu)\) and let \(X(\omega)=\omega\) for all \(\omega\in \mathbb{R}\). Then, \[ P(X\le x)= \mu(\{\omega: X(\omega)\le x \})= \mu(\{\omega: \omega\le x \})=\mu((-\infty,x])=F(x). \]
Second construction:
Let \((\Omega, \mathcal{F},P)=((0,1),\mathcal{B}(0,1),\lambda)\), where \(\lambda\) is Lebesgue measure on (0,1), and let \(X(\omega)=F^{-1}(\omega)\) for \(\omega\in (0,1)\). (Quantile function \(F^{-1}(\omega\)는 defined on \((0,1)\)이라 적절하다.) Then, \[ P(X\le x)= P(\{ \omega\in (0,1): F^{-1}(\omega)\le x \})= P(\{ \omega\in (0,1): \omega\le F(x) \})=F(x). \]