
Real Analysis Real Analysis Math Forum 
 LinkBack  Thread Tools  Display Modes 
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: 6,344 Thanks: 534 
Your definition seems in error. The second and third ranges overlap.


Tags 
$nu, $nu$, measure, show, signed 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Signed Up  gjb  New Users  4  April 14th, 2016 08:50 PM 
Show that the measure of X is different from zero.  walter r  Real Analysis  8  January 7th, 2014 07:09 PM 
show that [latex]\mu[/latex] is a measure  rayman  Real Analysis  2  October 14th, 2013 02:30 AM 
Signed hexadecimal computation!  eChung00  Computer Science  3  March 4th, 2013 02:59 PM 
what is a 8bit signed integer?  aaronmath  Computer Science  3  January 31st, 2013 05:07 AM 