Theorem. Suppose the random variable $U$ has a uniform $(0,1)$ distribution. Let $F$ be a continuous distribution function. Then the random variable $X=F^{-1}(U)$ has distribution function $F$.
In the text, the proof for the above theorem assumed that F was strictly monotone (so strictly increasing since the cdf cannot be decreasing, by definition). One of the exercises in the text is to strengthen the proof by assuming $F$ is monotone. To do this, they suggest to use the following definition for the inverse:
$$ F^{-1}(u) = \text{inf}\{x:F(x) \geq u\}, \quad 0<u<1 $$
A few comments/questions on this definition:
(1) I guess the reason for this new inverse definition is because we’re now assuming that $F$ is monotone and we can have a scenario such that for $x_1 \neq x_2$, $F^{-1}(u) = x_1$ and $F^{-1}(u) = x_2$, which would not make it a function. Am I correct to think this?
(2) Assuming the above is correct, I think the “greatest lower bound” part of the definition makes it so we only get one single value $x$ which makes the inequality true, specifically the “smallest” value. That being said, why do we require that $F(x) \geq u$ and not simply $F(x) = u$?
Help on this will be greatly appreciated.
(1) is correct.
(2) The reason for this is that a CDF can also be discontinuous (albeit it is always right-continuous by definition).
Consider a CDF $F(x)$ that jumps at $x=1/2$ from zero to one. Then, for $p\in(0,1)$ there is no $x$ such that $F(x)=p$ ever holds, but there is always an $x$ such that $F(x)\ge p$ holds.
Things to consider:
The generalized inverse $F^{-1}$, as correctly defined in OP and also called quantile function, is constant in some nonempty left neighbourhood $(q,p]$ of $p$ if and only if $F$ is discontinuous at $x=F^{-1}(p)$.
The CDF $F$ is constant in some nonempty left neighbourhood $(y,x]$ of $x$ if and only if $F^{-1}$ is discontinuous at $p=F(x)$.