How does the author of this probability book claim that a one-to-one function is invertible without showing it's onto? I know that the necessary conditions for a function to be invertible is that the function must be onto. I'm not sure how the author can just assume that the function $u(x)$ is onto.



Definition. Let $A$ and $B$ be sets. A function $f \colon A \to B$ is called invertible if it has an inverse, i.e. if there exists $g \colon B \to A$ such that $g \circ f = \operatorname{id}_A$ and $f \circ g = \operatorname{id}_B$.
Proposition. A function is invertible if and only if it is bijective.

Now:

Definition. Let $A$ and $B$ be sets. A function $f \colon A \to B$ is called left-invertible if it has a left-inverse, i.e. if there exists $g \colon B \to A$ such that $g \circ f = \operatorname{id}_A$.
Proposition. A function is left-invertible if and only if it is injective.

and similarly

Definition. Let $A$ and $B$ be sets. A function $f \colon A \to B$ is called right-invertible if it has a right-inverse, i.e. if there exists $g \colon B \to A$ such that $f \circ g = \operatorname{id}_B$.
Proposition. A function is right-invertible if and only if it is surjective.

Note that $\operatorname{id}_A$ is the identity function on $A$.
So basically the author of your book should have written "left-inverse" instead of just "inverse", but this is often omitted, especially since he also makes it clear by writing "so that $v(u(x))=x$", which means exactly $v \circ u = \operatorname{id}_{\mathbb R}$.
Bonus: to prove the last proposition you need the Axiom of Choice. Actually, the statement in the last proposition is equivalent to the Axiom of Choice.
See also: http://en.wikipedia.org/wiki/Inverse_function#Left_and_right_inverses
