Let X have a standard normal distribution and let $Y = 2X$. Determine whether (i) the cdf $F(x, y)$ of $(X,Y)$ is continuous, and (ii) the distribution of $(X, Y)$ is absolutely continuous with respect to the Lebesgue measure in the $(x, y)$ plane.
I'm not sure how to approach (i), I understand that the joint cdf of continuous random variables is
$$ \iint f(x,y) \,dx\,dy $$
but I'm not sure how to obtain $f(x, y)$ from the definitions of $X$ and $Y$.
On
I just posted this answer showing the c.d.f. is continuous.
This is not absolutely continuous with respect to Lebesgue measure on the plane, because the Lebesgue measure of the line $y=2x$ is $0,$ whereas this measure assigns the number $1$ to that set.
When you deal with continuous r.v. it is usually much more easy and clear to start working with their cdf, and that's actually what they ask for (not the density $f_{XY}$; in fact, you can't actually assume it exists).
Remember that the CDF is the real-valued function $F_{XY}$ that takes two real variables defined as $$F_{XY}(x,y)=P(X\le x \wedge Y\le y)$$ that in this case can be reduced to $$F_{XY}(x,y)=P(X\le x \wedge 2X\le y).$$
Then try to write an expression that relates this function to the (one variable) CDF of a normal standard r.v. (while there is not a nice formula for it you can just call it $\Phi(t)$ and all you need to know is that its derivative is the density function of $N(0,1)$ distribution.
Once you have a formula for $F_{XY}$ (which is not straightforward), you can see if it is a continuous function (this is equivalent to say thay $(X,Y)$ is a continuous r.v.) and if there is a function $f(x,y)\ge 0$ such that $$F_{XY}(x,y)=\iint \limits_{\{(s,t)\colon s\le x,\,t\le y\}} f(s,t)\; d\mu.$$
When this is the case, such an $f$ is called a 'probability density function' and we say that $(X,Y)$ is an absolutely continuous r.v.
Not an easy one, but is possible if you use patience and detail.