I am reading Rick Durrett's Probability: Theory and Examples ($5^{\text{th}}$ Edition) and the author states the following Theorem (Theorem $1.2.2$):
If $F:\mathbb{R}\rightarrow[0, \infty)$ satisfies the following:
$(i)$ $F$ is non-decreasing
$(ii)$ $\lim_{x\rightarrow\infty}F(x) =1$ and $\lim_{x\rightarrow -\infty}F(x) =0$
$(iii)$ $F$ is right-continuous i.e. $\lim_{y\downarrow x}F(y) = F(x)$
then $F$ is the distribution function of some random variable.
To prove the theorem, the author defines $\Omega = (0, 1)$, $\mathcal{F} = \mathcal{B}(0,1)$ and $P = \text{Lesbegue measure}$. The author then claims that $X(\omega):=\sup\{y\in \mathbb{R}:F(y)<\omega\}$ is the desired random variable.
Question: How does one prove that $X$ is measurable (this is not explicitly proved in the book)?
Thoughts: The author proves that
$$X^{-1}((-\infty, x]) = \{\omega\in\Omega: X(\omega)\leq x\} = \{\omega\in\Omega:\omega\leq F(x)\} = (-\infty, F(x)]\cap(0, 1)$$
From here one could claim that $X^{-1}(\sigma\{(-\infty, x]: x\in\mathbb{R}\}) = X^{-1}(\mathcal{B}(\mathbb{R})) = \sigma(\{(-\infty, F(x)]\cap(0,1): x\in\mathbb{R}\})$
But is it possible to prove that this $\sigma$-algebra is $\mathcal{B}(0,1)$?
Best Answer
Note that $X$ is a non-decreasing function. For any non-decreasing function $f$ observe that $\{x: f (x) <a\}$ is an interval for any real number $a$. Hence $X$ is measurable.