I am looking for feedback on my solution to the following problem: A random variable X has a uniform distribution on $(c, 4c)$ with $c > 0$. $Y$ is given by $ \left\{ \begin{array}{ll} x & x \in (2c,3c) \\ 0 & \text{OW} \end{array} \right.$
Find the distribution of $Y$.
$\textbf{My attempt:}$
We have that $S_X = \{ x: f_X(x)\geq 0 \} = (c, 4c) $ and $S_Y = \{y: y=g(x) \text{ for some x in } S_X\} = \{ y: y =g(x) \text{ for } x \in (2c,3c)\}$. Since $f$ is continuous on $S_X$ and $g^{-1}(y) = y$ has a continuous derivative on $S_Y$, we get: $$f_Y(y) = \left\{ \begin{array}{ll} f_X(g^{-1}(y))\left|\frac{dg^{-1}(y)}{dy}\right| & y \in (2c,3c) \\ 0 & \text{OW} \end{array} \right. = \left\{ \begin{array}{ll} \frac{1}{3c} & y \in (2c,3c) \\ 0 & \text{OW} \end{array} \right.$$
Thanks for any help/feedback you may have for me!
The resulting density is not absolutely continuous. so $f_Y(y)=\frac{2}{3}$ when $Y=0$
the rest is correct but no calculation is needed because in the interval you calculated it is given that $Y=X$
Now your density is good because you have
$$\frac{2}{3}+\int_{2c}^{3c}\frac{1}{3c}dy=1$$
So, concluding, you density is the following
$$f_Y(y) = \begin{cases} \frac{2}{3}, & \text{if $y=0$} \\ \frac{1}{3c}, & \text{if $2c<y<3c$} \\ 0, & \text{if OW} \end{cases}$$
EDIT: graph of the transformation function
as you can see, Y-domain is $\{0\} \cup (2c;3c)$