Proving Non-Negativity and Monotonicity of Integrals over a Measure Space

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SUMMARY

The discussion centers on proving the non-negativity and monotonicity of integrals over a measure space, specifically addressing the statement that if \( lf+-\varphi < \frac{\epsilon}{2\mu(E)} \), then \( \int_E lf+-\varphi < \frac{\epsilon}{2} \). The key conclusions are that if \( f \geq 0 \) almost everywhere, then \( \int f \, d\mu \geq 0 \), and if \( f \geq g \geq 0 \) on the set \( E \), then \( \int_E f \, d\mu \geq \int_E g \, d\mu \). These results are foundational in measure theory and integral calculus.

PREREQUISITES
  • Understanding of measure theory concepts, particularly measures and integrals.
  • Familiarity with simple functions and their properties.
  • Knowledge of the properties of non-negative functions in the context of integration.
  • Basic proficiency in mathematical notation and logic.
NEXT STEPS
  • Study the properties of Lebesgue integrals, focusing on non-negativity and monotonicity.
  • Explore the implications of the Dominated Convergence Theorem in measure theory.
  • Learn about the relationship between simple functions and measurable functions.
  • Investigate the concept of almost everywhere (a.e.) convergence in the context of integrals.
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Mathematicians, students of advanced calculus, and anyone studying measure theory or integration techniques will benefit from this discussion.

EV33
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Homework Statement


My question is would I be allowed to say,
if lf+-[itex]\phi[/itex]l<ε/(2[itex]\mu[/itex](E)
then ∫E lf+-[itex]\phi[/itex]l<ε/2

Homework Equations



E is the set in which we are integrating over.
[itex]\mu[/itex] is the measure
[itex]\varphi[/itex] is a simple function
f+ is the non-negative part of the function f.

The Attempt at a Solution



I can't think of any reason that this wouldn't be true but my text is very vague in this chapter and so I am really not sure if this is an ok statement.



Thank you for your time.
 
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Yes, your conclusion is correct.

It's not hard to prove the following:
##\newcommand{dmu}{\operatorname{d}\!\mu}##

(a) If ##f\geq 0## a.e., then ##\int f\dmu \geq 0##.
(b) If ##f\geq g\geq 0## on E, then ##\int_E f\dmu\geq\int_E g\dmu##.

Hint: To prove (b), use (a) and the fact that the assumption implies that ##f\chi_E\geq g\chi_E## (everywhere, and therefore a.e.).
 

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