Let $\delta_x$ denote the measure defined by
$$\delta_x(E)=\begin{cases}
1, & x\in E \\
0, & x\not\in E \\
\end{cases}
$$
Let $\mu:=m+\delta_0+\delta_1$, where $m$ denotes the Lebesgue measure on $\mathbb{R}$. Let $\lambda$ be a signed Borel measure on $\mathbb{R}$ such that for any continuously differentiable, compactly supported function $f:\mathbb{R}\to\mathbb{R}$,
$$\int_\mathbb{R} f(x)d\lambda(x)=\int_0^1f'(x)x^2dm(x)$$
I need to show that $\lambda$ is absolutely continuous with respect to $\mu$ and find its Radon-Nikodym derivative.
My attempt: If $\mu(E)=0$, then $m(E)=0$ and $0,1\not\in E$. I know I need to show that $\lambda(E)=0$. All I have is the following and I can't seem to proceed.
$$\lambda(E)=\int \chi_E d\lambda$$
Clearly $\chi_E$ is not continuously differentiable on $\mathbb{R}$. Where do I go from here? Hints are greatly appreciated. Thanks.
Best Answer
Hint: Use integration by parts on the right side. You can now guess what the R-N derivative should be. The answer is $\frac {d\lambda} {d\mu}=g$ where $g(x)=-2x$ for $0 <x<1$, $g(1)=1$ and $g(0)=0$. To show that this works you have to use the well known fact that if $\int fd\tau =\int f d\mu$ for all continuously differentiable $f$ with compact support and $\tau \{0\}=\mu \{0\}$, $\tau \{1\}=\mu \{1\}$ then $\tau =\mu$.
[Some details added in my comment below].