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Proof of a set is sigma finite

  1. Jan 23, 2013 #1
    1. The problem statement, all variables and given/known data

    if f is integrable, then the set N(f) = {x : f(x)≠ 0} is [itex]\sigma[/itex]-finite

    2. Relevant equations
    i am stucked in this proof , somebody help me please


    3. The attempt at a solution
    if f is simple the it seems the set is finite since otherwise the the integral won't exist but how can it be extended to f is integrable?
     
  2. jcsd
  3. Jan 23, 2013 #2

    Dick

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    What's the definition of N(f) being sigma finite? N(f) doesn't have to be finite. f(x)=1/x^2 is integrable on [1,infinity). [1,infinity) isn't finite.
     
    Last edited: Jan 23, 2013
  4. Jan 24, 2013 #3
    sorry i didn't put it right , it seems N(f) [itex]\sigma[/itex]-finite means the measure of N(f) is a countable union of finite measure sets
    u(N(f))= [itex]\underbrace{\cup}_{n}[/itex]u(N(fn)) which u(N(fn)) <[itex]\infty[/itex]
     
  5. Jan 24, 2013 #4

    Dick

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    Define a set A(f,n)={x: |f(x)|>1/n}. A(f,n) will have finite measure, right?
     
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