Math Problem Statement
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Solution
Let's go through the solution step-by-step:
Problem Statement
The problem is to show that if a function is integrable, then it must be finite-valued almost everywhere (a.e.).
Solution Breakdown
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Assumption: Assume for contradiction that is not finite-valued almost everywhere. This means that there exists a set with positive measure where for all .
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Implication of being infinite on : Since on the set , it implies that on . This leads us to consider the integral over the set , which would theoretically be infinite because the function takes an infinite value on a set of positive measure.
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Contradiction Setup: For any positive integer , consider the following inequality: Since on , this inequality will hold for any (no matter how large).
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Integrability Condition: However, if is integrable over the entire space, should be finite. But, our assumption that on with implies that is infinite, which contradicts the assumption that is integrable.
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Conclusion: Therefore, our assumption that takes infinite values on a set of positive measure must be false. Hence, is finite-valued almost everywhere, meaning that can only be infinite on a set of measure zero.
This concludes the proof.
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Related Questions
- What does it mean for a function to be integrable?
- How does the concept of measure relate to integrability?
- Can a function that takes on infinite values be integrable in certain contexts?
- What does "finite-valued almost everywhere" mean in the context of integration?
- Why is a contradiction a useful method in proofs like this one?
Tip:
When proving properties of integrable functions, it can be helpful to consider the behavior of the function over sets of positive measure and use inequalities to highlight contradictions with integrability assumptions.
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Math Problem Analysis
Mathematical Concepts
Real Analysis
Measure Theory
Lebesgue Integration
Formulas
∫ |f| dx > n * m(E)
Theorems
Contradiction Method in Analysis
Integrability Theorem
Suitable Grade Level
Graduate level
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