I have worked out the proof for this question but wanted to check if my intuition also holds here. Please see below for the brief proof and then my proposed explanation:
For a continuous random variable, $X$, with cdf, $F(x)$, then if we define a continuous random variable
$ U = F(X)$, then $U \thicksim U(0,1)$
For the proof, we need to show that $P(U \le u) = u$
$F(U) = P(U \le u)\\ = P(F(X) \le u)\\ = P\big(F^{-1}\left(F(X)\right) \le F^{-1}(u)\big)\\ = P\big(F^{-1}(U) \le F^{-1}(u) \big)\\ = F\big( F^{-1}(U)\big)\\ = u $
My question relates to whether this argument holds and I would appreciate if someone could explain if it doesnt:
$U \thicksim (0,1)$ and since (i) $u \in [0,1]$ and (ii) $F(U)$ is monotonic increasing, then $F(U) = P(U \le u) = u$
I am sure this probably isn't correct but would really appreciate if someone could point out where this does not hold as I rationalised this before solving the actual equation, so feel it was a more intuitive understanding.
Thanks a lot!
IF $F$ is strictly monotone increasing, the argument of the OP is correct. If not, then we have to consider the alternative to $F^{-1}$. This can be achieved by using the quantile function of $X$. $$ Q(t)=\inf\{x\in\mathbb{R}: F(x)\geq t\}\quad\text{where}\quad 0<t<1$$
It can be proven that $$ F(x)\geq t\qquad\text{if and only if}\qquad Q(t)\leq x$$ This equivalency implies that $F(Q(t))\geq t$ in general. However, if and that $F$ is continuous at $Q(t)$, then for any $s<Q(t)$, $F(s)<t$. Taking limit as $s\nearrow Q(t)$ gives $F(Q(t))=t$.
Since $F$ is continuous by assumption, $F(x)=\lim\limits_{y\nearrow x}F(y)=F(x-)$ for all $x\in\mathbb{R}$. Therefore, if $U=F(X)$ $$\begin{align} P[U<u]=P[F(X)<u]=P[X<Q(u)]=F(Q(u))=u \end{align}$$ This is enough to conclude that $U$ is uniform distributed over $(0,1)$.