# Solution Check-Finding The Radon Nikodym Derivative

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.

• its not true that $\mu(A) \le m(A)$ in general, so how do you justify the inequality? Eg $\mu((0,1))=\pi>1=m((0,1))$ Jul 14 at 5:10

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 $$a ].
Now consider $$\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$$.