I encountered the following example:
(Folded Normal). Let $Y = \vert Z \vert$ with $Z ∼ N(0, 1)$. The distribution of $Y$ is called a Folded Normal with parameters $\mu = 0$ and $\sigma^2 = 1$. At first sight, $Y$ may seem tricky to deal with since the absolute value function is not differentiable at $0$ (due to its sharp corner), but $Y$ has a perfectly valid continuous distribution.
$$E(Y) = E \vert Z \vert = \int_{-\infty}^\infty \vert z \vert \dfrac{1}{\sqrt{2\pi}} e^{-z^2/2} \ dz = 2 \int_0^\infty z \dfrac{1}{\sqrt{2\pi}} e^{-z^2/2} \ dz = \sqrt{\dfrac{2}{\pi}}.$$
How is it possible that, despite the fact that the absolute value function is not differentiable at $0$ (due to its sharp corner), $Y$ still has a continuous distribution? Is this not in violation of the theory behind integration in real analysis?
Thank you.