Suppose $X$ is standard Gaussian. The random variable $Y$ is $sgn(X)\sqrt{|X|}$. Is $Y$ absolutely continuous? The issue is obtaining a density. The problem is $sgn(X)$ since the derivative is undetermined around 0. But thinking about it, the singleton set ${0}$ still does have 0 measure since if we have the interval $(-\epsilon, 0]$, as $\epsilon \to 0$ then indeed the measure of our probability law goes to 0, following from the Gaussian, since $F_X(0) - F_X(-\epsilon) = \Phi(0) - \Phi(-\epsilon^2)$. Can I conclude anything? Besides 0, it would be possible to just chain rule and find a density function based on the Gaussian density function right? (although I'm not quite sure what the meaning of such a density function would be anymore)
I also was reading about how $0 \delta(0) = 0$, so at 0 the density is 0? That doesn't seem quite right. If anything I would expected a spike at 0. But again the singleton 0 seems to still have probability 0...
Having singletons be measure 0 only gives ordinary continuity.