$\{x_n\}$ is a sequence in $\mathbb{R}$ and $\{p_n\}$ is a sequence of positive numbers. Define a $\sigma$-finite measure $\nu(E)=\sum_{x_n\in E}p_n$. Find the Lebesgue decomposition of $\nu$ with respect to Lebesgue measure on $\mathbb{R}$.
My solution:
Let $S=\{x_n\}_{n\in\mathbb{N}}$. Assume $\lambda(E)=0,\forall E\subseteq \mathbb{R}$.
Define $\nu_1(E)=\nu(E\backslash S)=0$ because $E\backslash S$ does not have corresponding $\{p_n\}\Rightarrow \nu_1\ll\lambda$.
Define $\nu_2(E)=\nu(E\cap S)=\sum_{x_n\in E}p_n>0\Rightarrow \nu_2\perp\lambda$.
Any comment is appreciated.
