Why Does a Zero Integral Imply a Function Is Zero Almost Everywhere?

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SUMMARY

The discussion centers on the mathematical proof that if the integral of a measurable function \( f: X \to [0, \infty] \) over a measure space \( (X, m) \) is zero, then \( f \) must equal zero almost everywhere. The participant demonstrates that assuming \( f \) is positive on a set \( A \) with positive measure leads to a contradiction, as the integral over \( A \) would also be positive, contradicting the initial condition. This establishes that \( f \) cannot be non-zero on any set of positive measure.

PREREQUISITES
  • Understanding of measure theory concepts, particularly measure spaces.
  • Familiarity with the properties of integrals, especially in the context of non-negative functions.
  • Knowledge of measurable functions and their characteristics.
  • Basic proficiency in mathematical proof techniques, particularly proof by contradiction.
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  • Study the properties of Lebesgue integrals and their implications in measure theory.
  • Explore the concept of sets of measure zero and their significance in analysis.
  • Learn about the Dominated Convergence Theorem and its applications in integration.
  • Investigate further into the implications of the Lebesgue Dominated Convergence Theorem on function behavior.
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This discussion is beneficial for mathematics students, particularly those studying real analysis and measure theory, as well as educators seeking to clarify the implications of integrals in the context of measurable functions.

Ja4Coltrane
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Homework Statement



Let (X,m) be a measure space, f:X->[0, infinity] be measurable.

If the integral over X of f is 0, show that f=0 almost everywhere



Homework Equations





The Attempt at a Solution



Suppose that f is nonzero on A, m(A)>0.

I've reduced problem to proving that the following is impossible:

f is positive on a set of nonzero measure and
{x such that f(x)>epsilon} has zero measure for all epsilon greater than zero.

I don't know how to do that though...

Any help?
 
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Suppose the statement is not true- that there exist a set A, of measure greater than 0, on which f is positive. Then the integral of f over A alone is positive and, since f is never negative, the integral of f over X is not less than the integral of f over A.
 

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