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Equality of integrals => equality of integrands

  1. Jan 2, 2012 #1
    hi folks,

    one often reads

    [itex]\int_A f(x) dx = \int_A g(x) dx [/itex] for arbirary A, thus f(x) = g(x), since the equaltiy of the Integrals holds for any domain A.

    I don't see, why the argument "...for any domain A..." really justifies this conclusion.

    Can someone explain this to me, please?
     
  2. jcsd
  3. Jan 2, 2012 #2

    LCKurtz

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    For continuous functions and argument might go like this. If f and g aren't identical, there is a point a where one of them, say f(a), is greater than g(a). So there is an interval I containing a where f(x) > g(x). But then ##\int_I f(x)-g(x)\, dx > 0## contradicting the assumption "for any domain A".
     
  4. Jan 2, 2012 #3

    micromass

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    And for discontinuous functions, the result might not even be true!! It needs to be:

    If [itex]\int_A f(x)dx = \int_A g(x)dx[/itex] then [itex]f=g[/itex] almost everywhere.
     
  5. Jan 2, 2012 #4

    mathman

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    For the more general case the conclusion would be f(x)=g(x), except on a set of measure 0.
     
  6. Jan 2, 2012 #5
    I think you can do something like the following. Assuming the integrals of f and g are equal for every set A (and f and g are obviously measurable):

    Consider the function [itex] f(x) - g(x) [/itex]. Let E be the set of x's in A where this function is positive. Define

    [itex] E_{1/n} = \left{ x \in E : f(x) - g(x) > 1/n \right} [/itex]

    Thus by Tschebyshev,

    [itex] m(E_{1/n}) \leq n \int_{ E_{1/n} } f - g = 0 [/itex]

    Therefore

    [itex] m(E) = \cup_{n=1}^\infty m(E_{1/n}) = 0 [/itex]

    ie, the set where f is strictly greater than g has measure zero.

    You can do basically the same argument to show that the set where g is strictly greater than f has measure zero. So f and g are the same for ae x. It's not really a for any domain A type question, it just needs to be the case for these E_{1/n}'s.
     
    Last edited: Jan 2, 2012
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