Showing that two random variables defined in different probability spaces have the same distribution

Question

Let the distribution functions of $X_n$ be $F_n$ and assume that $F_n(x)$ is a continuous and strictly increasing function in $x$. Note that this implies that the inverse function of $F_n$ exists and is strictly increasing and continuous on $(0,1)$. Let $\Omega' = (0,1)$ be the unit interval and $F'$ be the Borel $\sigma$-algebra on $\Omega'$ and $P'$ be the uniform measure on $\Omega'$. Let $Y_0, Y_1, ... $ be random variables defined on $\{\Omega', F', P'\}$ such that $Y_n(\omega') = F_n^{-1}(\omega')$ for $n \in \{0,1,2,...\}$ and $\omega' \in \Omega$.

Show that for $n \in \{0,1,2,...\}$ the random variable $Y_n$ has the same distribution as $X_n$.

I don't really know how to interpret that $Y_n(\omega') = F_n^{-1}(\omega')$, where does this random variables map to? By definition I know that $F_{X_n} = P(\{\omega \in \Omega; X_n(\omega) \leq x\})$ How can I define the distribution function of $Y_n$?

I would appreciate any advice.

Answer 1

So $Y_n$ would be defined on $(0,1)$ and take values in $(0,1)$ since $F_n$ maps into $(0,1)$. As for its distribution, $P'(Y_n\leq t)$, we compute: $$P'(Y_n\leq t) =P'(\{\omega : \omega \leq F_n(t)\})=F_n(t),$$ by definition of $P'$.