Integral over a set of measure 0

  • Context: Undergrad 
  • Thread starter Thread starter Boot20
  • Start date Start date
  • Tags Tags
    Integral Measure Set
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
7 replies · 10K views
Boot20
Messages
8
Reaction score
0
Is the integral over a set of measure zero always equals to zero? Can the integral be undefined?
 
Physics news on Phys.org
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 [tex]\infty \cdot 0 = 0[/tex], so, in this case, even the function takes [tex]\infty[/tex] 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 [itex]+\infty[/itex] or [itex]-\infty[/itex].
 
Last edited:
But apparently to me [tex]\infty \cdot 0 = 0[/tex] should be adopted
Else, if [tex]f[/tex] admit [tex]\infty[/tex] on set [tex]A[/tex] of measure [tex]0[/tex], we may use [tex]f_{n} = n[/tex] on [tex]A[/tex] to approximate [tex]f[/tex] from below, then, the integral of [tex]f_{n}[/tex] is zero, by monotone convergence theorem, the integral of [tex]f[/tex] should be zero as well. If we do not define [tex]\infty \cdot 0 = 0[/tex], we may get inconsistency in this case?
 
wayneckm said:
But apparently to me [tex]\infty \cdot 0 = 0[/tex] should be adopted
Else, if [tex]f[/tex] admit [tex]\infty[/tex] on set [tex]A[/tex] of measure [tex]0[/tex], we may use [tex]f_{n} = n[/tex] on [tex]A[/tex] to approximate [tex]f[/tex] from below, then, the integral of [tex]f_{n}[/tex] is zero, by monotone convergence theorem, the integral of [tex]f[/tex] should be zero as well. If we do not define [tex]\infty \cdot 0 = 0[/tex], 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 [tex]f[/tex] to be integrated?