Absolute continuity of a distribution function This appeared on an exam I took.
$Z \sim \text{Uniform}[0, 2\pi]$, and $X  = \cos Z$ and $Y = \sin Z$. Let $F_{XY}$ denote the joint distribution function of $X$ and $Y$.
Calculate $\mathbb{P}\left[X+ Y \leq 1\right]$. So this was easy - 
$$\begin{align}
 \mathbb{P}\left[X+Y \leq 1\right] &= \mathbb{P}\left[\sin Z+ \cos Z \leq 1\right] \\
&=\mathbb{P}\left[\sqrt{2}\sin\left(Z+\frac{\pi}{4}\right)\leq 1\right] \\
&= \mathbb{P}\left[Z \leq \arcsin\frac{1}{\sqrt{2}} - \frac{\pi}{4} \right] \\
&= \dfrac{\arcsin\frac{1}{\sqrt{2}} - \frac{\pi}{4}}{2\pi}
\end{align}
$$
But then, the question asked if $F_{XY}$ was absolutely continuous. I don't think so, but how would I prove it?
I thought about proceeding like this
$$ 
\begin{align}
F_{XY}(x, y) &= \mathbb{P}\left[X \leq x, Y \leq y\right],\; x, y \in [0, 1] \\
&= \mathbb{P}\left[Z \leq \min(\arccos x, \arcsin y)\right]
\end{align}
$$
This is definitely continuous, but is it absolutely continuous?
Thanks!

Absolutely continuous distributions are measures with a density with respect to the Lebesgue measure. The distribution of $(X,Y)$ is concentrated on the unit circle, which has Lebesgue measure zero, hence this distribution is not absolutely continuous.
Regarding the computation of $\mathrm P(X+Y\leqslant1)$, note that $\arcsin(1/\sqrt2)=\pi/4$ hence your formula would yield $\mathrm P(X+Y\leqslant1)=0$ (a quite unlikely result). Rather, drawing a picture of the unit circle and of the line of equation $x+y=1$ shows that $x+y\leqslant1$ corresponds to the three quarters of the circle from the point $(0,1)$ to the point $(1,0)$ through the points $(-1,0)$ and $(0,-1)$, hence $\mathrm P(X+Y\leqslant1)=3/4$.
