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Integrands that vanish everywhere

  1. Jan 24, 2006 #1


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    In physics, we often use the assumption that if the integral:

    [tex]\int_D f(\vec x) d \vec x =0[/tex]

    and this holds for any region D in the space, then the integrand must vanish identically everywhere in the space. This was motivated by the argument that if it didn't vanish in some region, we could focus the integrand in that region (say, only where it is positive) and the above equation wouldn't hold. But I was thinking, wouldn't this argument only show that the integrand vanishes almost everywhere. That is, it could still be non-zero on a set of measure zero, and the above equation would still always be satisfied. Is it just that this possibility doesn't concern physicists, or is there actually a way to show it must vanish at every point?
  2. jcsd
  3. Jan 24, 2006 #2


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    Actually, now I see that the argument would work if the function were continuous, because then if there was any point x where f(x) were non-zero, say positive, it would have a neighborhood where f was positive, and so we could integrate over that neighborhood to get a non-zero value. That's good enough for me. You can delete this thread if you want.
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