Prove $X(\omega) = \sup\{y \in \mathbb{R}: F(y) < \omega\}$ is a random variable.

Solution 1:

If $X\left(\omega\right)\leq x<z$ then if follows immediately from the definition $X\left(\omega\right):=\sup\left\{ y\mid F\left(y\right)<\omega\right\} $ that $F\left(z\right)\geq\omega$.

The fact that $F$ is continuous on the right then allows the conclusion that also $F\left(x\right)=\lim_{z\rightarrow x+}F\left(z\right)\geq\omega$.

In conversely $F\left(x\right)\geq\omega$ then $X\left(\omega\right)=\sup\left\{ y\mid F\left(y\right)<\omega\right\} \leq x$ because $F$ is non-decreasing.

Proved is now that $$X\left(\omega\right)\leq x\iff\omega\leq F\left(x\right)$$

This equation gives us the second part:

$X:\Omega\rightarrow\mathbb{R}$ is a measurable function and this with: $$F_{X}\left(x\right)=P\left\{ \omega\mid X\left(\omega\right)\leq x\right\} =\lambda\left\{ \omega\in\left(0,1\right)\mid\omega\leq F\left(x\right)\right\} =\lambda\left((0,F\left(x\right)]\right)=F\left(x\right)$$