Proof of a set is sigma finite

manuel huant
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Homework Statement



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

Homework Equations


i am stucked in this proof , somebody help me please

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?
 
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manuel huant said:

Homework Statement



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

Homework Equations


i am stucked in this proof , somebody help me please

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?

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.
 
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sorry i didn't put it right , it seems N(f) \sigma-finite means the measure of N(f) is a countable union of finite measure sets
u(N(f))= \underbrace{\cup}_{n}u(N(fn)) which u(N(fn)) <\infty
 
manuel huant said:
sorry i didn't put it right , it seems N(f) \sigma-finite means the measure of N(f) is a countable union of finite measure sets
u(N(f))= \underbrace{\cup}_{n}u(N(fn)) which u(N(fn)) <\infty

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