I was hoping to get my solution to part $\textbf{i}$ of this qual question regarding the Radon-Nikodym derivative checked for rigor and correctness. Then I was hoping to get advice on proceeding with part $\textbf{ii}$. Here is the question:
Let $ m$ be the Lebesgue measure and define a measure $ \mu$ on the Borel $\sigma$-algebra on $[0,1]$ by the formula $$\mu(X) = m(\{y\in [0,\pi]:\sin(y) \in X\}) .$$ $\textbf{i}$ Show that $\mu$ is absolutely continuous with respect to the Lesbesgue measure.
$\textbf{ii.}$ Find the Radon Nikodym derivative $\frac{d\mu}{dm}. $
Here is my solution:
$\textbf{i}$ Let $A\subset \mathbb{R} $ such that $m(A) = 0$. We have that $$ \mu(A) = m(\{y\in [0,\pi]:\sin(y) \in A\}) $$ and from monotonicity we have $$\mu(A) \leq m(A\cap [0,\pi]) \leq m(A) = 0$$ and therefore $\mu << m $.
$\textbf{ii}$ Since $m $ is a $\sigma$-finite positive measure and $\mu $ is a finite positive measure (notice that $\mu(\mathbb{R}) = \pi)$ and $\mu << m $ by the Radon-Nikodym theorem there exists a $m$-integrable non-negative function $f $ which is measurable with respect to the Borel $\sigma $-algebra and $$\mu(A) = \int_A f d\mu $$
Furthermore, if any other function $g $ satisfies above, then $f=g $ a.e.
Notice that the set $B=[0,1] $ has full measure with respect to $\mu $. I want to claim that $$ f= 2\sin^{-1}(x) \cdot \chi_{[0,1]}$$ is the Radon Nikodym derivative, but I'm not exactly sure how to show this.
Your inequality is not correct. Here are some hints:
Let $\mu_1(X)=m\{x \in [0,\frac {\pi} 2]: \sin y \in X\}$. Verify that the derivative of $\sin^{-1} x$, namely $\frac 1 {\sqrt {1-x^{2}}}$ is the RND of $\mu_1$ w.r.t. $m$. [ For this show that $\int_a^{b} \frac 1 {\sqrt {1-x^{2}}} dx=\mu_1 (a,b)$ for $a <b$].
Now consider $\mu_2(X)=m\{x \in [\frac {\pi} 2, \pi]: \sin y \in X\}$. Use the fact that $\sin (\pi -y) =\sin y$ to prove absolute continuity of $\mu_2$ and for writing down the RND of $\mu_2$.