I try to prove:
Let $X$ be a discrete random variable with cdf $F$, $F_{-}(x):=P(X<x)$, $W \sim U[0,1]$ a random variable and $X, W$ independent.
Then $$V:=WF(X)+(1-W)F_{-}(X) \sim U[0,1].$$
My ideas: $$F_V(v|X=x)=P(V\leq v|X=x)=\frac{P(\lbrace V\leq v\rbrace \cap \lbrace X=x \rbrace)}{P(X=x)}$$ and since $$V\leq v \Leftrightarrow WF(X)+(1-W)F_{-}(X) \leq v \Leftrightarrow W \leq \frac{v-F_{-}(X)}{F(X)-F_{-}(X)}$$ we get $$F_V(v|X=x)=\frac{P(\lbrace W \leq \frac{v-F_{-}(X)}{F(X)-F_{-}(X)}\rbrace \cap \lbrace X=x \rbrace)}{P(X=x)}.$$ Because $X,W$ independent: $$F_V(v|X=x)=\frac{P(W \leq \frac{v-F_{-}(X)}{F(X)-F_{-}(X)})P( X=x )}{P(X=x)}.$$ And since $W\sim U[0,1]$ we finally get $$F_V(v|X=x)=\frac{v-F_{-}(X)}{F(X)-F_{-}(X)}=\frac{v-F_{-}(X)}{P(X=x)}.$$
We can conclude: $$F_V(v)=P(V\leq v)=\sum\limits_x P(V\leq v |X=x)P(X=x)=\sum\limits_x v-F_{-}(x)$$
So my question now is: What's wrong about this? And how can I deduct that $V \sim U[0,1]$?