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August 25th, 2017, 02:07 PM  #1 
Newbie Joined: Aug 2017 From: uganda Posts: 1 Thanks: 0  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. 
August 26th, 2017, 01:59 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,497 Thanks: 580 
Your definition seems in error. The second and third ranges overlap.


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$nu, $nu$, measure, show, signed 
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