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$\{x_n\}$ 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.

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You cannot assume $\lambda(E) = 0 \quad \forall\, E \subseteq \mathbb{R}$. I suppose that is not what you wanted to write. –  Thomas Dec 10 '12 at 9:15
    
Thank you, @Thomas. –  Sam Dec 10 '12 at 15:20

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