Undergrad Integral over a set of measure 0

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SUMMARY

The integral over a set of measure zero is definitively zero for ordinary functions, according to Lebesgue integration principles. This conclusion is supported by the convention that \(\infty \cdot 0 = 0\), which prevents inconsistencies when dealing with functions that may take on infinite values. Riemann integration, however, does not apply in cases involving extended-real-number-valued functions. The monotone convergence theorem further reinforces that if a function approximated by a sequence of functions \(f_n\) has an integral of zero, then the integral of the original function \(f\) must also be zero.

PREREQUISITES
  • Understanding of Lebesgue integration
  • Familiarity with the monotone convergence theorem
  • Knowledge of Riemann integration principles
  • Concept of measure theory, specifically measure zero sets
NEXT STEPS
  • Study the properties of Lebesgue integration and its implications for measure theory
  • Explore the monotone convergence theorem in detail
  • Investigate the differences between Lebesgue and Riemann integration
  • Learn about functions defined on sets of measure zero and their integrals
USEFUL FOR

Mathematicians, students of real analysis, and anyone interested in advanced calculus concepts, particularly in the context of integration and measure theory.

Boot20
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Is the integral over a set of measure zero always equals to zero? Can the integral be undefined?
 
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It may be undefined if the function itself is peculiar with infinity as its value. For ordinary functions the integral will be 0.
 
Last edited:
mathman said:
It may be undefined if the function itself is peculiar with infinity as its value. For ordinary functions the integral will be 0.

Sorry, I have the question that, if Lebesgue integration, they always define the convention \infty \cdot 0 = 0, so, in this case, even the function takes \infty in a set of measure 0, the integral is still 0?
 
Convention is an easy way out.
 
For Lebesgue integration, that the integral over a set of measure zero is a rather trivial theorem, following from the fact that all simple functions have integral zero -- so via (what I believe is) the usual formulation, it doesn't even need to be treated as a special case.


Riemann integration assumes the function is real-valued, so it doesn't even apply if you are considering extended-real-number-valued functions that take on the values +\infty or -\infty.
 
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But apparently to me \infty \cdot 0 = 0 should be adopted
Else, if f admit \infty on set A of measure 0, we may use f_{n} = n on A to approximate f from below, then, the integral of f_{n} is zero, by monotone convergence theorem, the integral of f should be zero as well. If we do not define \infty \cdot 0 = 0, we may get inconsistency in this case?
 
wayneckm said:
But apparently to me \infty \cdot 0 = 0 should be adopted
Else, if f admit \infty on set A of measure 0, we may use f_{n} = n on A to approximate f from below, then, the integral of f_{n} is zero, by monotone convergence theorem, the integral of f should be zero as well. If we do not define \infty \cdot 0 = 0, we may get inconsistency in this case?

But what if the f_{n} are defined over sets of measure non-zero, but that the sum of the measure of those sets converges to zero?
 
Last edited:
Boot20 said:
But what if the f_{n} are defined over sets of measure noe intgen-zero, but that the sum of the measure of those sets converges to zero?

Sorry that I cannot get your idea exactly. Or can you tell me explicitly what is the function f to be integrated?
 

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