Let F be a distribution function. On $(\Omega, \mathfrak{F}, P)=((0,1), \mathfrak{B}(0,1),\lambda)$ where $\lambda$ denotes Lebesgue measure.
Define $X: \Omega \to \mathbb{R}$ by $X(\omega) = \sup\{y \in \mathbb{R}: F(y) < \omega\}$.
1 Show that $\forall x \in \mathbb{R}, (\omega: X(\omega) \leq x) = (\omega: \omega \leq F(x))$
2 Show that X is a RV in $(\Omega, \mathfrak{F}, P)$ and that $F_X = F$.
1 LHS = $(\omega: X(\omega) \leq x)$
= $(\omega: X(\omega) \in (-\infty, x])$
= $(\omega: \sup(y \in \mathbb{R}: F(y) < \omega) \in (\infty, x])$
RHS = $(\omega: \omega \leq F(x))$
= $(\omega: \omega \in (-\infty, F(x)])$
= $(\omega: \omega \in (-\infty, P(X^{-1}( \ (-\infty,x] \ ))))$
= $(\omega: \omega \in (-\infty, P(LHS)))$
I'm stuck. Help please? :(
Cross posted : https://stats.stackexchange.com/questions/110704/let-f-be-a-distribution-function-prove-that-x-is-a-rv
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: $$F_{X}\left(x\right)=P\left\{ \omega\mid X\left(\omega\right)\leq x\right\} =\lambda\left\{ \omega\in\left(0,1\right)\mid\omega\leq F\left(x\right)\right\} =\lambda\left((0,F\left(x\right)]\right)=F\left(x\right)$$