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
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 pleaseThe 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...
i got this question when i read the proof of stone-weierstrass theorem in baby rudin , page 159 , this inequality seems right when n becomes larger, since 1-nx^2 would be negative and (1-x^2)^n always positive, but i don't know how to proved it rigorously using binomial theorem for all n , or is...