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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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Hint: Consider $f_1 = f|_{(-\infty,0]}$ and $f_2 = f|_{(0,\infty)}$ and write $f = f_1 + f_2$. – t.b. Apr 12 '12 at 2:56
@t.b. thanks for the hint. please can I get further hints. thanks. – Jay Apr 12 '12 at 4:55
Put $\nu_1(E)=\int_{E}f_1\,d\lambda$ and $\nu_2(E)=\int_{E}f_2\,d\lambda$. Notice that $\nu = \nu_1+\nu_2$, $\nu_1 \ll \mu$ and $\nu_2 \perp \mu$. The point is that the support of $g$ divides the support of $f$ into two pieces of non-zero measure, $(-\infty,0]$ and $[0,1]$. – t.b. Apr 12 '12 at 10:19
@t.b. Would I be correct in setting $A=(-\infty,0],~B=(0,\infty)$. then $v_1(E) = v(E\cap A)$ and $v_2(E) = v(E\cap B)$? – Jay Apr 12 '12 at 14:59
Yes, this amounts to the same. – t.b. Apr 12 '12 at 15:01

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