probability theory - Prove that F: $mathbb{R} to mathbb{R}$ nondecreasing, right continous and goes to 1/0 in $infty/-infty$ is a CDF for a random variable X

Prove that F: $\mathbb{R} \to \mathbb{R}$ nondecreasing, right continous, $\lim_{t \to -\infty} F(t) = 0$, $\lim_{t \to \infty} F(t) = 1$ is a CDF for some random variable X, i.e. there exists random variable $X$ such that $F_X = F$ where $F_X$ denotes CDF of X.

The hint was to look onto $\left(\Omega=(0,1), \mathcal{F}=\mathcal{B}((0,1)), \lambda_{|(0,1)}\right)$.

I know that CDF of any random variable has those properties, but I fail to see how to link $F_X$ with $F$.

Answer

For $\omega \in (0,1)$ define $X(\omega)=\inf \{t: F(t) \geq \omega\}$. First note that $F(t) \geq \omega\}$ implies that $X(\omega) \leq t$.

Now suppose $F(t) <\omega$. Then $F(s) <\omega$ for all $s \leq t$. This implies that $X(\omega) \geq t$. In other words $X(\omega) implies $F(t) \geq \omega$.

Thus $P(X\leq t) \leq\lambda((0,F(t)])=F(t)$ or $F_X(t) \leq F(t)$. Also $F_X(t-) \leq F(t)$. These two inequalities imply that $F_X(t)=F(t)$ at all points where $F_X$ is continuous. Since both of these are right-continuous function it follows that $F_X=F$.