Probability integral transformation : Why doesn't this inverse affect inequality? I am reading this problem from DeGroot's "Probability & Statistics" 2nd edition.


I can't understand why $$\Pr[G^{-1}[F(X)]\leq z]=\Pr[F(X)\leq G(z)]$$, as nowhere is strict monotonicity assumed. I don't get why $G^{-1}[F(X)]\leq z\iff F(X)\leq G(z)$.
 A: You are correct that $G$ might not be strictly increasing, and thus the equivalence $G^{-1}[F(X)]\leq z\iff F(X)\leq G(z)$ cannot be deduced. However, it holds in the probabilistic sense the probability of one of these conditions being true and another false is 0. 
Let us decompose the event $F(X)\leq G(z)$ into two cases depending on whether $G^{-1}[F(X)]\leq z$:
\begin{equation}
P( F(X) \leq G(z) ) = P\left([F(X)\leq G(z)] \cap [G^{-1}[F(X)]\leq z]\right) + P\left([F(X)\leq G(z)] \cap [G^{-1}[F(X)] > z] \right).
\end{equation}
Due to (not necessarily strict) monotonicity of $G$, we have $G^{-1}[F(X)]\leq z \Rightarrow F(X) \leq G(z)$. This implies that the first term of the decomposition is actually $P(G^{-1}[F(X)]\leq z)$. It remains to be shown that the second term is 0.
Note that by monotonicity of $G$, $F(X)<G(z)$ implies $G^{-1}[F(X)]\leq z$ and thus the event in question, $[F(X)\leq G(z)] \cap [G^{-1}[F(X)] > z]$,  is a subset of $F(X)=G(z)$. Hence,
\begin{equation}
P\left([F(X)\leq G(z)] \cap [G^{-1}[F(X)] > z] \right)\leq P(F(X)=G(z)).
\end{equation}
But, $X$ was assumed to have a continuous d.f., for which it was shown that $F(X) \sim U(0,1)$. In particular, the probability of $F(X)$ being equal to any particular value is 0. Thus, 
\begin{equation}
P\left([F(X)\leq G(z)] \cap [G^{-1}[F(X)] > z] \right) = 0.
\end{equation}
Substituting this into the second term of the first equation, we conclude that
\begin{equation}
P\left( F(X) \leq G(z) \right) = P\left(G^{-1}[F(X)]\leq z\right).
\end{equation}
