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August 25th, 2017, 02:07 PM   #1
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Show that $\nu$ is a signed measure

Let
$$f(x)=
\begin{cases}
x& \text{if}~x\geq 1;\\
0& \text{if}~-1<x\leq 1; \\
-x^2& \text{if}~x\leq 1;\\
\end{cases}
$$
Given that $d\nu = fd\mu$ is a measure. Show that $\nu$ is a signed measure, where $\mu$ is the usual Lebesgue measure and $f$ is an integrable function.
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August 26th, 2017, 01:59 PM   #2
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Your definition seems in error. The second and third ranges overlap.
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