If $X$ has a density, then does $Y:=g(X)\cdot 1_{\{X>0\}}$? (No...?) I think there is a typo in my probability theory book's exercises. Probability Essentials by Jacod and Protter, exercise 11.9 says: 

Let $X$ have a density, and let $$Y=ce^{-\alpha X}1_{\{X>0\}}, \qquad (\alpha > 0, c>0).$$ Find $f_Y(y)$ in terms of $f_X.$ [Ans: $f_Y(y)=\frac{f_X\left(-\frac{1}{\alpha}\ln\left(\frac{y}{c}\right)\right)}{\alpha y}1_{(0,c)}(y).$]

I don't think $Y$ necessarily has a density. $F_Y(y)=\int 1_{\{ce^{-\alpha X}1_{\{X>0\}}\leq y\}} dP$. If $X$ takes nonpositive values with nonzero probability, won't $F_Y$ be discontinuous at 0? 

I believe you are right. If $E\sim\mathrm{Exp}(1)$ is an exponential random variable, then choosing $X=-E$ yields a random variable with a density whereas $Y=0$ a.s. does not have a density.
One can show in general that if $\mathbb P(X\le0)>0$ then $Y$ does not have a density.
