The Problem: Suppose that $X$ is absolutely continuous with a PDF $$f_X(t)=\begin{cases}e^{-t}&\text{if }t\geq0\\0&\text{if }t<0.\end{cases}$$ Let $$g(x)=\begin{cases}x&\text{if }x\leq5\\6&\text{if }x>5.\end{cases}$$ Find the CDF of $Y=g(X)$. Check whether this is an absolutely continuous random variable and find its PDF if it has one.
My Thoughts: First we find the CDF of $X$ for later use. We have $$F_X(t)=\int_{-\infty}^tf_X(s)\,ds=\begin{cases}0&\text{if }t<0\\1-e^{-t}&\text{if }t\geq0.\end{cases}$$ Therefore, $P(X\leq0)=0$. Next, if $0<t<6$ then $g(X)\leq t$ if and only if $X\leq t$ so that $$P(g(X)\leq t)=P(X\leq t)=1-e^{-t}.$$ It follows that $$F_Y(t)=\begin{cases}0&\text{if }t\leq0\\1-e^{-t}&\text{if }0<t<6\\1&\text{if }t\geq6.\end{cases}$$ Note that $F_Y$ is not continuous at $t=6$ and hence $Y$ is not absolutely continuous. Moreover, $Y$ cannot have a PDF since $\displaystyle\int_{-\infty}^\infty\frac{d}{dt}F_Y(t)\,dt\ne1.$
My Difficulties: For some reason I am not convinced of my work, I think I made a mistake somewhere because the answer seems too easy. Could someone please confirm or deny my worries? Thank you for your time and highly appreciate any feedback.