A question about lebesgue integral

  • Context: Graduate 
  • Thread starter Thread starter pswongaa
  • Start date Start date
  • Tags Tags
    Integral
Click For Summary
SUMMARY

The discussion centers on the properties of the Lebesgue integral, specifically addressing whether the condition ∫f² dμ = 0 implies that f = 0 almost everywhere (a.e.) on the interval. The conclusion is that this is not necessarily true, as counterexamples exist where the positive and negative parts of f² cancel each other out. The participants clarify that for any measurable function g on a measure space (X, Σ, μ), the condition ∫|g| dμ = 0 indeed implies g = 0 a.e., but this must be contextualized within the nature of the function (real vs. complex).

PREREQUISITES
  • Understanding of Lebesgue integration
  • Familiarity with measure theory concepts
  • Knowledge of measurable functions
  • Basic properties of real and complex functions
NEXT STEPS
  • Study the properties of Lebesgue integrals in detail
  • Explore counterexamples in measure theory
  • Learn about the implications of integrals equaling zero in different contexts
  • Investigate the differences between real and complex functions in integration
USEFUL FOR

Mathematicians, students of analysis, and anyone studying measure theory and Lebesgue integration who seeks to deepen their understanding of function properties in integration contexts.

pswongaa
Messages
7
Reaction score
0
if lebesgue integral of f^2 over an interval equal 0, must f=0 a.e on that interval?
 
Physics news on Phys.org
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.
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 33 ·
2
Replies
33
Views
7K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 20 ·
Replies
20
Views
4K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 16 ·
Replies
16
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K