A question about lebesgue integral

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  • #1
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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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  • #2
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No. Try to find a counterexample (hint: the integral can be 0 since positive and negative parts cancel out).
 
  • #3
D H
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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.


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?
 
  • #4
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Oh god. Never mind my reply.
 
  • #5
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For any measure space [itex](X,\mathcal{S},μ)[/itex], and any measurable function [itex]g:\rightarrow [-∞,∞][/itex], [tex]∫|g|dμ=0\implies g=0 a.e.[/tex]

Specifically, since [itex]f^2=|f^2|[/itex], this gives [itex]f^2=0[/itex] a.e., and hence [itex]f=0[/itex] a.e.
 
  • #6
D H
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You are assuming f is a real function, Axiomer. If it's a complex function, then f2 is not the same as |f2|.
 
  • #7
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That's a good point. Since the op didn't specify otherwise, I assumed we were talking about functions to the extended real line.
 
  • #8
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For any measure space [itex](X,\mathcal{S},μ)[/itex], and any measurable function [itex]g:\rightarrow [-∞,∞][/itex], [tex]∫|g|dμ=0\implies g=0 a.e.[/tex]

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

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

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

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