I am learning measure theory through self-studying and am trying to solve an exercise related to the Radon-Nikodym derivative. Specifically, Exercise 1.4.3 on page 57 in The Theory of Statistics and Its Applications, which asks:
Given a random variable $X = \max\{x_0, Y\}$ with $x_0 > 0$ and $Y$ being another r.v. with Lebesgue density $f_Y(y) = e^{-y}$ for $y > 0$ and $f_Y(y)=0$ otherwise. Show that $X$ has a Lebesgue density $f_X$ w.r.t. the measure $\mu = m + \delta_{x_0}$ ($m$ being the Lebesgue measure and $\delta_{x_0}$ is the unit point mass measure) and that $$ f_X(x) = \frac{d(P\circ X^{-1})}{d\mu}(x) = \begin{cases} e^{-x} \qquad x > x_0\\ 1-e^{-x_0} \qquad x = x_0\\ 0 \qquad \text{otherwise} \end{cases}$$ where $P$ is the probability measure of an underlying probability space $(\Omega, \mathcal{F}, P)$.
I am struggling to show that $f_X(x) = e^{-x}$ if $x > x_0$. More concretely, I see that $P(X = x > x_0) = P(Y = y > x_0)$, but fail to use this information for the desired result (see below, but think it should probably be trivial). (I have shown that $P \circ X^{-1} << \mu$ and that $(\mathbb{R}, \mathcal{B}, \mu)$ is $\sigma$-finite (with $\mathcal{B}$ referring to the Borel $\sigma$-field). Furthermore, I have shown the two cases $f_X(x) = 1 - e^{-x_0}$ for $x = x_0$ and $f_X(x) = 0$ for $x < x_0$.)
What I have done for the case $x > x_0$:
Assume we are given any $A \in \mathcal{B}$ with $x > x_0$ for all $x \in A$. Then $\{\omega \in \Omega | X(\omega) \in A\} = \{\omega \in \Omega | Y(\omega) \in A\}$ and hence, $P_X(A) = P_Y(A)$ (using notation $P_X(A) = P\circ X^{-1}(A)$). Thus, as we know the Lebesgue density of $Y$, using the Radon-Nikodym Theorem gives us
$$ P \circ X^{-1}(A) = \int_A f_X d\mu = \int_A f_Y d\mu = P \circ Y^{-1}(A) $$
I do think that $f_X = f_Y$ probably (easily) follows from the uniqueness of the Radon-Nikodym derivative. However, I cannot see how. In particular, I understand the uniqueness as if $\int_A f_X d\mu = \int_A f_Y d\mu$ would be valid for all $A \in \mathcal{B}$, we would have $f_X = f_Y$ $\mu$-a.e. However, we look at a restricted set of measurable sets only having elements $x > x_0$. So what if there is another function $f \ne f_Y$ only happening to have the same Lebesgue integral value for this restricted set of measurable sets? How can I show this is not the case?
Furthermore, if I would use the uniqueness property, I would probably end up with a statement that $f_X = f_Y$ $\mu$-a.e. I.e., $f_X$ is not determined on $\mu$-null sets. To get to the solution, would I choose a specific form or version of $f_X$ for $\mu$-null sets, as it does not change the integral? Here in particular, $f_X = f_Y$?
Note: There are multiple questions on StackExchange asking about concrete Radon-Nikodym derivates. However, none is directly applicable to this case and hence, I opened a new question.