Proof of a set is sigma finite

[manuel huant]

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?

 

[Dick]

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.

 

[manuel huant]

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)) <$\infty$

 

[Dick]

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?