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Following mpiktas answer, things became clearer to me and I think I can know answer the question. The first statement

$F_U(U(\bar{\omega})) = F_X(z_{\bar{w}})$ for all $\bar{\omega} \in \Omega \implies z_{\bar{\omega}}=X(\bar{\omega})$

Is in fact equivalent to (and better understood as)

$F_U(U({\omega})) \sim F_X(Z(\omega)) \implies Z(\omega)=X({\omega})$, where $\sim$ means "distributed as".

Now, this turns out not to be quite true (see mpiktas comment below). What is true however is

$F_U(U({\omega})) \sim F_X(Z(\omega)) \implies Z(\omega) \sim X({\omega})$.

As noticed by mpiktas, $F_U(U({\omega})) \sim U(0,1)$ ( a proof can be found here). Hence we must have $F_X(Z(\omega)) \sim U(0,1)$. But then we must have $Z(\omega) \sim X(\omega)$. All that is needed for the proof of this last fact can also be found here. But in the link, the statement is reversed ($F_X(X(\omega)) \sim U(0,1)$ instead of $F_X(Z(\omega)) \sim U(0,1) \Rightarrow Z(\omega) \sim X(\omega)$ ) so I figured it might be helpful to reformulate it here.

• Assume $F_X(Z(\omega)) \sim U(0,1)$.
• Then $F_{F_X(Z(\omega))}(a) = a$
• Notice that \begin{align} a &= F_X(F^{-1}_X(a))\\ &= P(X\leq F^{-1}_X(a))\\ &= P(F_X(X) \leq a)\\ &= F_{ F_X(X)} (a) \end{align} where I omitted the $\omega$'s for notational convenience.
• So we have $F_{F_X(Z(\omega))}(a) = a = F_{ F_X(X(\omega))} (a)$, which in turn implies $Z(\omega) \sim X(\omega)$.

Following mpiktas answer, things became clearer to me and I think I can know answer the question. The first statement

$F_U(U(\bar{\omega})) = F_X(z_{\bar{w}})$ for all $\bar{\omega} \in \Omega \implies z_{\bar{\omega}}=X(\bar{\omega})$

Is in fact equivalent to (and better understood as)

$F_U(U({\omega})) \sim F_X(Z(\omega)) \implies Z(\omega)=X({\omega})$, where $\sim$ means "distributed as".

Now, this turns out not to be quite true (see mpiktas comment below). What is true however is

$F_U(U({\omega})) \sim F_X(Z(\omega)) \implies Z(\omega) =^d X({\omega})$, where $=^d$ means that the two random variables have the same distribution.

As noticed by mpiktas, $F_U(U({\omega})) \sim U(0,1)$ ( a proof can be found here). Hence we must have $F_X(Z(\omega)) \sim U(0,1)$. But then we must have $Z(\omega) =^d X(\omega)$. All that is needed for the proof of this last fact can also be found here. But in the link, the statement is reversed ($F_X(X(\omega)) \sim U(0,1)$ instead of $F_X(Z(\omega)) \sim U(0,1) \Rightarrow Z(\omega) =^d X(\omega)$ ) so I figured it might be helpful to reformulate it here.

• Assume $F_X(Z(\omega)) \sim U(0,1)$.
• Then $F_{F_X(Z(\omega))}(a) = a$
• Notice that \begin{align} a &= F_X(F^{-1}_X(a))\\ &= P(X\leq F^{-1}_X(a))\\ &= P(F_X(X) \leq a)\\ &= F_{ F_X(X)} (a) \end{align} where I omitted the $\omega$'s for notational convenience.
• So we have $F_{F_X(Z(\omega))}(a) = a = F_{ F_X(X(\omega))} (a)$, which in turn implies $Z(\omega) =^d X(\omega)$.

Following mpiktas answer, things became clearer to me and I think I can know answer the question. The first statement

$F_U(U(\bar{\omega})) = F_X(z_{\bar{w}})$ for all $\bar{\omega} \in \Omega \implies z_{\bar{\omega}}=X(\bar{\omega})$

Is in fact equivalent to (and better understood as)

$F_U(U({\omega})) \sim F_X(Z(\omega)) \implies Z(\omega)=X({\omega})$, where $\sim$ means "distributed as".

Now, as noticed by mpiktas, $F_U(U({\omega})) \sim U(0,1)$ ( a proof can be found here). Hence we must have $F_X(Z(\omega)) \sim U(0,1)$. But then we must have $Z(\omega) = X(\omega)$. All that is needed for the proof of this last fact can also be found here. But in the link, the statement is reversed ($F_X(X(\omega)) \sim U(0,1)$ instead of $F_X(Z(\omega)) \sim U(0,1) \Rightarrow Z(\omega) = X(\omega)$ ) so I figured it might be helpful to reformulate it here.

• Assume $F_X(Z(\omega)) \sim U(0,1)$.
• Then $F_{F_X(Z(\omega))}(a) = a$
• Notice that \begin{align} a &= F_X(F^{-1}_X(a))\\ &= P(X\leq F^{-1}_X(a))\\ &= P(F_X(X) \leq a)\\ &= F_{ F_X(X)} (a) \end{align} where I omitted the $\omega$'s for notational convenience.
• So we have $F_{F_X(Z(\omega))}(a) = a = F_{ F_X(X(\omega))} (a)$, which in turn implies $Z(\omega) = X(\omega)$ (not completely sure about this last step...could it be that is does not hold?).

$Z(\omega), X(\omega)$ are functions so by $Z(\omega) = X(\omega)$ we really mean $Z(\bar{\omega}) = X(\bar{\omega})$ for all $\bar{\omega} \in \Omega$, which connects this answer to the initial statement of the question.

Following mpiktas answer, things became clearer to me and I think I can know answer the question. The first statement

$F_U(U(\bar{\omega})) = F_X(z_{\bar{w}})$ for all $\bar{\omega} \in \Omega \implies z_{\bar{\omega}}=X(\bar{\omega})$

Is in fact equivalent to (and better understood as)

$F_U(U({\omega})) \sim F_X(Z(\omega)) \implies Z(\omega)=X({\omega})$, where $\sim$ means "distributed as".

Now, this turns out not to be quite true (see mpiktas comment below). What is true however is

$F_U(U({\omega})) \sim F_X(Z(\omega)) \implies Z(\omega) \sim X({\omega})$.

As noticed by mpiktas, $F_U(U({\omega})) \sim U(0,1)$ ( a proof can be found here). Hence we must have $F_X(Z(\omega)) \sim U(0,1)$. But then we must have $Z(\omega) \sim X(\omega)$. All that is needed for the proof of this last fact can also be found here. But in the link, the statement is reversed ($F_X(X(\omega)) \sim U(0,1)$ instead of $F_X(Z(\omega)) \sim U(0,1) \Rightarrow Z(\omega) \sim X(\omega)$ ) so I figured it might be helpful to reformulate it here.

• Assume $F_X(Z(\omega)) \sim U(0,1)$.
• Then $F_{F_X(Z(\omega))}(a) = a$
• Notice that \begin{align} a &= F_X(F^{-1}_X(a))\\ &= P(X\leq F^{-1}_X(a))\\ &= P(F_X(X) \leq a)\\ &= F_{ F_X(X)} (a) \end{align} where I omitted the $\omega$'s for notational convenience.
• So we have $F_{F_X(Z(\omega))}(a) = a = F_{ F_X(X(\omega))} (a)$, which in turn implies $Z(\omega) \sim X(\omega)$.

Following mpiktas answer, things became clearer to me and I think I can know answer the question. The first statement

$F_U(U(\bar{\omega})) = F_X(z_{\bar{w}})$ for all $\bar{\omega} \in \Omega \implies z_{\bar{\omega}}=X(\bar{\omega})$

Is in fact equivalent to (and better understood as)

$F_U(U({\omega})) \sim F_X(Z(\omega)) \implies Z(\omega)=X({\omega})$, where $\sim$ means "distributed as".

Now, as noticed by mpiktas, $F_U(U({\omega})) \sim U(0,1)$ ( a proof can be found here). Hence we must have $F_X(Z(\omega)) \sim U(0,1)$. But then we must have $Z(\omega) = X(\omega)$. All that is needed for the proof of this last fact can also be found here. But in the link, the statement is reversed ($F_X(X(\omega)) \sim U(0,1)$ instead of $F_X(Z(\omega)) \sim U(0,1) \Rightarrow Z(\omega) = X(\omega)$ ) so I figured it might be helpful to reformulate it here.

• Assume $F_X(Z(\omega)) \sim U(0,1)$.
• Then $F_{F_X(Z(\omega))}(a) = a$
• Notice that \begin{align} a &= F_X(F^{-1}_X(a))\\ &= P(X\leq F^{-1}_X(a))\\ &= P(F_X(X) \leq a)\\ &= F_{ F_X(X)} (a) \end{align} where I omitted the $\omega$'s for notational convenience.
• So we have $F_{F_X(Z(\omega))}(a) = a = F_{ F_X(X(\omega))} (a)$, which in turn implies $Z(\omega) = X(\omega)$.

$Z(\omega), X(\omega)$ are functions so by $Z(\omega) = X(\omega)$ we really mean $Z(\bar{\omega}) = X(\bar{\omega})$ for all $\bar{\omega} \in \Omega$, which connects this answer to the initial statement of the question.

Following mpiktas answer, things became clearer to me and I think I can know answer the question. The first statement

$F_U(U(\bar{\omega})) = F_X(z_{\bar{w}})$ for all $\bar{\omega} \in \Omega \implies z_{\bar{\omega}}=X(\bar{\omega})$

Is in fact equivalent to (and better understood as)

$F_U(U({\omega})) \sim F_X(Z(\omega)) \implies Z(\omega)=X({\omega})$, where $\sim$ means "distributed as".

Now, as noticed by mpiktas, $F_U(U({\omega})) \sim U(0,1)$ ( a proof can be found here). Hence we must have $F_X(Z(\omega)) \sim U(0,1)$. But then we must have $Z(\omega) = X(\omega)$. All that is needed for the proof of this last fact can also be found here. But in the link, the statement is reversed ($F_X(X(\omega)) \sim U(0,1)$ instead of $F_X(Z(\omega)) \sim U(0,1) \Rightarrow Z(\omega) = X(\omega)$ ) so I figured it might be helpful to reformulate it here.

• Assume $F_X(Z(\omega)) \sim U(0,1)$.
• Then $F_{F_X(Z(\omega))}(a) = a$
• Notice that \begin{align} a &= F_X(F^{-1}_X(a))\\ &= P(X\leq F^{-1}_X(a))\\ &= P(F_X(X) \leq a)\\ &= F_{ F_X(X)} (a) \end{align} where I omitted the $\omega$'s for notational convenience.
• So we have $F_{F_X(Z(\omega))}(a) = a = F_{ F_X(X(\omega))} (a)$, which in turn implies $Z(\omega) = X(\omega)$ (not completely sure about this last step...could it be that is does not hold?).

$Z(\omega), X(\omega)$ are functions so by $Z(\omega) = X(\omega)$ we really mean $Z(\bar{\omega}) = X(\bar{\omega})$ for all $\bar{\omega} \in \Omega$, which connects this answer to the initial statement of the question.