Prove that the two random variables $X$ and $Y$ with $X \sim \text{Unif}(0, 1)$ and $Y = X$ have no joint probability density function (PDF), while each margin has a PDF.
Here is my progress, please help me grade it and complete the proof!
Note that if $B = \{ (x, y) \in (0, 1) \times (0, 1) : x = y \}$, then $$1 = \underset{B}{\int\limits \int\limits} f_{X, Y}(x, y) = \int\limits_{x=0}^1 \quad \int\limits_{y=x}^x f_{X, Y}(x, y) = 0$$ a contradiction. Grade this solution, please. I do not have a clue how to do the second part.
The joint cdf of $X$ and $Y$ are given below for any $(x, y) \in (0, 1)^2$:
$$F_{X,Y}(x,y)=\mathbb P(X\le x,Y\le y)=P(X\le \min(x,y))= \min(x,y)$$
You can see that there is no $f_{X,Y}(u,w)$ such that for any $(x, y) \in (0, 1)^2$ the following representation holds:
$$F_{X,Y}(x,y)=\int\limits_{u=0}^x \quad \int\limits_{w=0}^y f_{X, Y}(u, w) \text{d}u\text{d}w.$$
Indeed, if such a function exits, then it (almost everywhere) satisfies
$$f_{X, Y}(x, y)=\frac{\partial^2{F_{X, Y}(x, y)}}{\partial{x}\partial{y}},$$
but the RHS is zero on non-diagonal points $(x,y)$ with $x \ne y$ and not defined on diagonal $(x,x)$ points. Hence, for a zero function $f_{X,Y}(u,w)=0$ , the above integral representation cannot be given.
In fact, $X$ and $Y$ have a continuous distribution, without a joint density function, while its margins have density functions.