Let $F$ be a probability distribution function on $\Bbb R$. For each $ω∈(0,1)$ , define $X(ω)=\inf\{x∈R:F(x)>ω\}$.
(a) Show that $X$ defines a real-valued function on $(0, 1)$, which is non-decreasing and hence measurable with respect to the Borel σ-field on $(0, 1)$.
(b) Show that for any real $x$,the Lebesgue measure of the set $\{ω ∈ (0, 1) : X(ω) ≤ x\}$ equals $F(x)$.
(c) Taking $Ω=(0,1)$, $A = \beta(0, 1)$ and $P$ = Lebesgue measure on $(0,1)$, conclude that,given any probability distribution function $F$ on $\Bbb R$, we have a random variable $X$ on $(Ω, A, P)$, whose distribution function is $F$ .
My attempt:
I did part (a) as a routine computation.
(b) $\lambda\{\omega \in (0,1):X(\omega) \le x\}=\text{inf }\mu(A \supseteq \{\omega \in (0,1):X(\omega) \le x\})$ $$=\text{inf }\mu(A \supseteq \{\omega \in (0,1):\inf \{z \in \Bbb R: F(z) > \omega\} \le x\})$$
How to simplify this to obtain $F(x)$ ?
(c) It's apparent that doing part (a) and (b) gives us the construction for $X$ on $(Ω, A, P)$, whose distribution function is $F$ .
Thanks in Advance for help!
For $\omega\in\left(0,1\right)$ let $Y\left(\omega\right):=\inf\left(\left\{ x\in\mathbb{R}\mid F\left(x\right)\geq\omega\right\} \right)$.
On base of this it is immediate that $F\left(x\right)\geq\omega\implies x\geq Y\left(\omega\right)$.
Fortunately the opposite direction is also true.
This because we have $F\left(x\right)\geq\omega$ for every $x>Y\left(\omega\right)$ so that the fact that $F$ is right- continuous implies that also $F\left(Y\left(\omega\right)\right)\geq\omega$.
So for every $\omega\in\left(0,1\right)$ and every $x\in\mathbb{R}$ we have: $$F\left(x\right)\geq\omega\iff x\geq Y\left(\omega\right)\tag1$$
From $\left(1\right)$ it follows that: $$\lambda\left(\left\{ \omega\in\left(0,1\right)\mid Y\left(\omega\right)\leq x\right\} \right)=\lambda\left(\left\{ \omega\in\left(0,1\right)\mid\omega\leq F\left(x\right)\right\} \right)=\lambda\left(\left(0,F\left(x\right)\right]\right)=F\left(x\right)\tag2$$
But we were asked to prove that for $X$, so we are not ready yet.
Now observe that $Y\leq X$ because $\left\{ x\in\mathbb{R}\mid F\left(x\right)>\omega\right\} \subseteq\left\{ x\in\mathbb{R}\mid F\left(x\right)\geq\omega\right\} $.
Consequently $\left\{ \omega\in\left(0,1\right)\mid X\left(\omega\right)\leq x\right\} \subseteq\left\{ \omega\in\left(0,1\right)\mid Y\left(\omega\right)\leq x\right\} $ for every $x\in\mathbb{R}$.
Now for some $\omega\in\left(0,1\right)$ let it be that $X\left(\omega\right)>x$ and $Y\left(\omega\right)\leq x$.
Here $X\left(\omega\right)>x$ implies that $F\left(x\right)\leq\omega$, and $Y\left(\omega\right)\leq x$ implies that $F\left(x\right)\geq\omega$.
Then $\omega=F\left(x\right)$.
Conversely we also have $F(x)\in\left\{ \omega\in\left(0,1\right)\mid Y\left(\omega\right)\leq x\right\}$ on base of $(1)$ and proved is now that: $$\left\{ \omega\in\left(0,1\right)\mid Y\left(\omega\right)\leq x\right\} =\left\{ F\left(x\right)\right\} \cup\left\{ \omega\in\left(0,1\right)\mid X\left(\omega\right)\leq x\right\}\tag3$$
Then based on $(2), (3)$ and the fact that singleton $\{F(x)\}$ has measure $0$ we conclude that:$$\lambda\left(\left\{ \omega\in\left(0,1\right)\mid X\left(\omega\right)\leq x\right\} \right)=F\left(x\right)\tag4$$