Let $X(x,y)=x$, and $Y(x,y)=y$, and let $P$ be a probability measure (on an appropriate space) that distributes mass uniformly over the boundary of the square defines by vertices $(0,0), (0,1),(1,0),(1,1)$.
What would be the joint $PDF$ of $X,Y$?
What is the CDF of $Y$?
For the first question, since the mass is distributed uniformly the joint PDF should be a constant, than integrates to one over the boundary. I'm not quite sure how to write this out, but my attempt is $$ 2\int_{0}^1cdx + 2\int_{0}^1cdy = 1 \implies c= \frac14 $$ which makes intuitive sense to me.
I'm not sure how to find the CDF of $Y$, though, as normally I would integrate the joint PDF to get the marginal pdf of Y, then integrate this over the relevant region w.r.t $Y$. However, integrating over the square seems wrong since the distribution is not uniform over the square, only the boundary.
My best guess is that $$ P(Y\leq y) = \frac14 + 2\int_0^y \frac14 dy \quad 0\leq y <1 $$ (and then the obvious cases when $y\leq 0$ and $y\geq 1$
If this is correct, is there a nicer way to do it (or a way where I can integrate over the whole square and not just the boundary?)
Decided to expand comment to answer.
You are right to approach this over the boundary, not the two-dimensional plane since the probability is concentrated there. Since the probability is squeezed into one dimension, the joint distribution of $(X,Y)$ is singular. So it has no joint PDF. We can still write an expression in terms of delta functions: $$ f_{X,Y}(x,y) = \frac{1}{4}\big(\delta(x-0)+\delta(x-1)+ \delta(y-0)+\delta(y-1)\big) $$ where $0\le x,y\le 1.$ Each of these four pieces represents one side of the square and the delta functions enforce that the density is only on the square boundary.
$Y$ should be a mixture of half uniformly distributed on $[0,1]$, one quarter atom at $0$ and one quarter atom at $1.$ So your answer for $Y$ looks right.