Requirements for transformation functions in probability theory

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I have two questions about transformation functions in probability theory. I wrote my answers to them but I'm not sure if I am right.

Let's look at the theorem below (I took it from here (the last)).

Theorem. Let X\displaystyle X be an absolutely continuous random variable with probability density function fX(x)\displaystyle f_X(x) and support RX={xR:f(x)>0}\displaystyle R_X = \{x \in \mathbb{R}: f(x) > 0 \}. Let g:RR\displaystyle g: \mathbb{R} \to \mathbb{R} be one-to-one and differentiable on the support of X\displaystyle X. If

dg1(y)dy0,yg(RX)\displaystyle \frac{dg^{-1}(y)}{dy} \ne 0, \qquad \forall y \in g(R_X)

then the probability density of Y\displaystyle Y is

fY(y)=fX(g1(y))dg1(y)dy,yg(RX).\displaystyle f_Y(y) = f_X(g^{-1}(y)) \left|{\frac{dg^{-1}(y)}{dy}}\right|, \qquad \forall y \in g(R_X).

Questions.

1) Can I choose any open subset A, RXAR\displaystyle A, ~R_X \subseteq A \subseteq \mathbb{R} which contains support RX\displaystyle R_X as the domain of g\displaystyle g in the thorem?

As I understood R\displaystyle \mathbb{R} is a "natural domain" (i.e. the largest possible domain) for g\displaystyle g but we can always limit it for a specific function g\displaystyle g on practice. I think set A\displaystyle A must be open to ensure the existence of the derivative of g\displaystyle g in all points of RX\displaystyle R_X (set RX\displaystyle R_X can be closed).

For example it will be convenient to choose A=(0,)\displaystyle A = (0,\infty) as the domain for a function g:AR, g(x)=ln(x)\displaystyle g:A \to \mathbb{R},~ g(x) = \ln(x) if X\displaystyle X can have only positive values.
Or I have to choose R\displaystyle \mathbb{R} as the domain and somehow define function g\displaystyle g on (,0]\displaystyle (-\infty,0] ?

2) Why the theorem requires function g\displaystyle g to be one-to-one instead of to be invertible?

I think that the domain of a function g1\displaystyle g^{-1} is exactly the set RY=g(RX)\displaystyle R_Y = g(R_X). Hence if g\displaystyle g is one-to-one then g\displaystyle g will be invertible function. Even if g\displaystyle g is formally not an onto function (not all points in codomaing=R\displaystyle \operatorname{codomain} g = \mathbb{R} have pre-images).


Please correct me if I'm wrong. Thanks.
 
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