Let $X$ have uniform distribution on the interval $(0, 1)$. Given $X = x$, let $Y$ have uniform distribution on the interval $(0, x)$.
Find the probability that $Y \le a ,\ 0 <a < 1$
I have found the joint density $f(x,y) = 1/x$ but I'm not sure how to continue.
By the law of total probability we have that $$P(Y\le a) = \int_0^1P(Y\le a|X=x)f_X(x)dx. \tag{1}$$
Since we know that $(Y|X=x) \sim U(0,x),$ we have that $$P(Y\le a | X=x) = \begin{cases}0, & a <0, \\ a/x, & 0<a<x,\\ 1, &a>x,\end{cases}$$ and we know that $$f_X(x) = \frac{1}{1-0} = 1.$$ Thus, we have $$(1) = \int_0^a1dx + \int_a^1\frac{a}{x}dx = a + a\ln(1/a) = a(1+\ln(1/a))$$