I want to find the distribution of $$Y_n(\omega) = \sin(2\pi n \omega)$$ on the probability space $(]0,1[, \mathcal B(]0,1[), \lambda|_{]0,1[})$ for $n\in \mathbb N$.
Let $X$ ~ $\text{Unif}_{]0,1[}. $We have $$f_{Y_n}(y) = f_X\left (\frac{\arcsin(y)}{2 \pi n} \right)\frac{d}{dy}\frac{\arcsin(y)}{2 \pi n} = \mathbb 1_{]0,1[} \frac{1}{2 \pi n \sqrt{1-y²}}$$ but integrating that over $]0,1[$ gives $\frac{1}{4n}$ and not $1$. What am I doing wrong?
$\newcommand{\P}{\mathbb P}$You are considering a probability space with the lebesgue measure. Then you can find the CDF "manually". The CDF is: $$\P(Y_n\leq z)= \frac{\arcsin(z)}{\pi}+\frac{1}{2}$$ How? Let $z\geq 0$. We first write $$\P(Y_n\leq z)=\P(Y_n<0)+\P(0\leq Y_n\leq z) $$ Finding $\{0\leq Y_n\leq z\}$ is finding $\omega\in (0,1)$ such that $0\leq\sin(2\pi n \omega)\leq z$. You have all numbers $\omega$ such that $0<2\pi n \omega \leq \arcsin(z)$. That means $\omega \in (0, \arcsin(z)/(2\pi n)]$. But a simple reasoning tells us that we have $2n$ intervals such that the inequality holds and they all have the same measure as $\lambda( (0, \arcsin(z)/(2\pi n)])$. That gave us $\P(0< Y_n\leq z) $. Now notice that $Y_n$ is negative with probability $1/2$, that gives us $\P(Y_n<0) =1/2$. That gives us the desired result.
With a similar reasoning, we can get the CDF for $z<0$ which I leave it up to you. What might help is the properties of the sine functions and drawing the sine will almost surely help.