Lebesgue Integration: Finite Measure Not Sufficient

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SUMMARY

The discussion centers on the limitations of finite measure in Lebesgue integration, specifically addressing the scenario where a measurable space (S, ℳ, μ) with μ(S) < ∞ does not guarantee that the integral of a non-negative function f: S → [0, ∞) is finite. The example provided involves the measurable space S = (0, 1) and the function f(x) = 1/x², demonstrating that the integral ∫₋₁¹ (1/x²) * (1/(1+x²)) dx diverges to infinity. This confirms that finite measure alone is insufficient for ensuring a finite integral.

PREREQUISITES
  • Understanding of Lebesgue integration
  • Familiarity with measurable spaces and measures
  • Knowledge of improper integrals
  • Basic calculus, particularly integration techniques
NEXT STEPS
  • Study the properties of Lebesgue measurable functions
  • Explore the concept of σ-finite measures
  • Learn about convergence theorems in Lebesgue integration
  • Investigate examples of functions that lead to divergent integrals
USEFUL FOR

Mathematicians, students of real analysis, and anyone studying measure theory and integration techniques will benefit from this discussion.

wayneckm
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Hello all,

Here is my question:

Suppose a measureable space [tex](S,\mathcal{S},\mu)[/tex] with [tex]\mu(S) < \infty[/tex] and [tex]f : S \mapsto [0,\infty)[/tex], this is not yet sufficient to ensure [tex]\int_{S} f d \mu < \infty[/tex].

Am I correct?
 
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S=(0,1) and f=1/x
 
Yes. Consider the measure defined on R by

[tex]\mu(E) =\int_E \frac 1 {1+x^2}\ dx[/tex]

for Lebesgue measurable E. Let f(x) = 1/x2. Then

[tex]\int_R \frac 1 {x^2}\cdot \frac 1 {1+x^2}\ dx \ge \int_{-1}^1 \frac 1 {x^2}<br /> \cdot \frac 1 2\ dx =\infty[/tex]
 

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