A uniform distribution falling between CDF values has the same distribution

Question

Let $F$ be any CDF and $U\sim Unif(0,1)$. Define $Y=y$ if $F(y-1)<U<F(y)$. I have seen the claim, and just proved it on my own, that if $F$ is integer-valued then $Y$ has CDF equal to $F$. That's pretty cool and interesting, it feels like it has something to do with the fact that $F_X^{-1}(X)\sim U$. So it's making me wonder, is this result also true if we drop the assumption that $F$ is integer-valued? I have a hard time seeing how we would extend the proof since my proof made use of a summation where terms cancelled, which doesn't seem available to a continuous version of the same problem. Perhaps you integrate and divide by ... something? Take a limit as $a\rightarrow 0^+$?

Answer 1

Since $\ Y\le y \iff U < F(y)\ $, then $$ \text{Prob}\left(Y\le y\right) = \text{Prob}\left(U\lt F(y)\right) = F(y)\ . $$