Let $B=\left\{b\right\}$ denote the set of atoms of the distribution function G. Define the quantile function $G^{-1}\left( a\right) = inf \left\{ x \in R : G(x) \ge a \right\}$.
Let V be independent of $X$ and uniformly distributed on [0,1]. Set $$\ U=\left\{ \begin{array}{cc} G\left( X\right) & if~X\not\in B \\ G\left( b-\right) +G\left\{ b\right\} V & if~X=b\in B% \end{array}% \right. , $$ where $G\left\{ b\right\} = G\left( b\right) - G\left( b-\right)$. Then, $U$ is uniformly distributed on $\left[ 0,1\right] $ and $% X=G^{-1}\left( U\right)$.
I am having a hard time proving this result. Can anyone provide any help?
Note that if $X=b\in B$ and $b<G^{-1}(u)$, then $$ U=G(b^-)+G\{b\}V\le G(b)<u. $$ Hence $U\ge u$ implies that $b\ge G^{-1}(u)=:w$. For each $u\in (0,1)$, we will calculate $P(U\ge u)$ dividing it into two cases.
Firstly, if it holds $G\{w\}=0$, then $X\in B$ and $X=b\ge w$ implies $U\ge G(b^-)\ge G(w^-)=G(w)=u.$ Also by the first proposition, $U\ge u$ implies $b\ge w$. If $X\notin B$ then $U=G(X)\ge u$ if and only if $X\ge w$. Hence we have $$\begin{eqnarray} P(U\ge u)&=&P(U\ge u, X\notin B)+P(U\ge u,X\in B)\\ &=&P(X\ge w, X\notin B) +P(X\ge w, X\in B)=P(X\ge w)=1-G(w^-)=1-u. \end{eqnarray}$$
Secondly, assume $G\{w\}>0$ and $U\ge u$. In case $X\notin B$, $U=G(X)\ge u$ is equivalent to $X\ge w$. In case $X\in B$, we see that $X=b>w$ implies $U\ge G(b^-)\ge G(w)\ge u$. A more subtle case is when $X=w\in B$. Then $U=G(w^-)+G\{w\}V\ge u$ if and only if $V\ge \frac{u-G(w^-)}{G\{w\}}$. Summing this up, we have $$\begin{eqnarray} P(U\ge u)&=&P(U\ge u, X\notin B)+P(U\ge u,X\in B)\\ &=&P(X\ge w, X\notin B) +P(X> w, X\in B)+P(X= w, V\ge \frac{u-G(w^-)}{G\{w\}})\\ &=&P(X\ge w, X\notin B) +P(X> w, X\in B)+P(X= w) \left(1-\frac{u-G(w^-)}{G\{w\}}\right)\\ &=&P(X>w) + G\{w\}\left(1-\frac{u-G(w^-)}{G\{w\}}\right)\\ &=&1-G(w)+G\{w\}-u+G(w^-)=1-u. \end{eqnarray}$$
Thus, it follows that $U$ is uniformly distributed on $[0,1]$.