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For a measurable set, $E$, let $\mu(E) = \int_E g~d\lambda, v(E) = \int_E f~d\lambda$ where $ f(x) = \begin{cases} \sqrt {1-x} & x\leq 1 \\ 0 & x\gt 1\end{cases} ,~~~ g(x) = \begin{cases} x^2 & x\leq 0 \\ 0 & x\gt 0\end{cases}.$ I want find the Lebesgue decomposition of $v$ with respect to $\mu$.

So I must find measures $v_1$ and $v_2$ such that $v_1$ is absolutely continuous with respect to $\mu$ and $v_1$ and $\mu$ are mutually singular. Also, I must have $v= v_1 + v_2.$ How do I find such $v_1$ and $v_2$. Is there a general technique? If it helps me, I know that $v + \mu$ is absolutely continuous with respect to $\lambda$.

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    Yes, this amounts to the same.2012-04-12

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