A question about lebesgue integral

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if lebesgue integral of f^2 over an interval equal 0, must f=0 a.e on that interval?
 
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No. Try to find a counterexample (hint: the integral can be 0 since positive and negative parts cancel out).
 
R136a1 said:
No. Try to find a counterexample (hint: the integral can be 0 since positive and negative parts cancel out).
What negative parts, R136a1? He's integrating f(x)2 over some interval.
pswongaa said:
if lebesgue integral of f^2 over an interval equal 0, must f=0 a.e on that interval?
Is f a function that maps reals to reals, or something else?
 
Oh god. Never mind my reply.
 
For any measure space (X,\mathcal{S},μ), and any measurable function g:\rightarrow [-∞,∞], ∫|g|dμ=0\implies g=0 a.e.

Specifically, since f^2=|f^2|, this gives f^2=0 a.e., and hence f=0 a.e.
 
You are assuming f is a real function, Axiomer. If it's a complex function, then f2 is not the same as |f2|.
 
That's a good point. Since the op didn't specify otherwise, I assumed we were talking about functions to the extended real line.
 
For any measure space (X,\mathcal{S},μ), and any measurable function g:\rightarrow [-∞,∞], ∫|g|dμ=0\implies g=0 a.e.

proof:
Define A=\{x\in X: g(x)≠0\}. For all naturals n, define A_n=\{x\in X: |g(x)|>\frac{1}{n}\}.

\frac{1}{n}μ(A_n)=∫\frac{1}{n}x_{A_n}dμ≤∫|g|dμ=0, so μ(A_n)=0 for all n.

Then μ(A)=μ(\bigcup _{n=1}^∞A_n)≤\sum _{n=1}^∞μ(A_n)=0\implies μ(A)=0, as desired.
 
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