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Measure theory question on integrals.

  1. Jun 12, 2013 #1
    Hi, I was wondering whether if ∫f×g dμ=∫h×g dμ for all integrable functions g implies that f = h?

  2. jcsd
  3. Jun 12, 2013 #2


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    f = h almost everywhere, assuming the integrals exist.
    Last edited: Jun 12, 2013
  4. Jun 12, 2013 #3


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    equality of integrals never implies equality, only equality a.e.
  5. Jun 13, 2013 #4
    Hi, thanks for the replies. However is this something trivial or is it hard to prove (that thery're equal a.e.)?
  6. Jun 13, 2013 #5
    Your claim was not properly explained, but I believe it is trivial to reduce it to this claim:

    If [itex]\int\limits_X f(x)d\mu(x)=0[/itex] and [itex]f(x)\geq 0[/itex] for all [itex]x\in X[/itex], then [itex]f(x)=0[/itex] for [itex]\mu[/itex]-"almost all" [itex]x\in X[/itex].

    This claim is not trivial. You must use the properties of measures and integrals.

    Assume that there exists a set [itex]A\subset X[/itex] such that [itex]f(x)>0[/itex] for all [itex]x\in A[/itex], and also [itex]\mu(A)>0[/itex]. Now you can define sets

    A_1 = \{x\in A\;|\; f(x)>1\}
    A_n = \Big\{x\in A\;\Big|\; \frac{1}{n-1}\geq f(x) > \frac{1}{n}\Big\},\quad\quad n=2,3,4,\ldots

    Equality [tex]\mu(A)=\sum_{n=1}^{\infty}\mu(A_n)[/tex] will imply that at least one of the [itex]\mu(A_n)[/itex] is positive.
    Last edited: Jun 13, 2013
  7. Jun 13, 2013 #6

    \int\limits_X \big(f(x)-h(x))g(x)d\mu(x)=0

    holds for all integrable [itex]g[/itex], then it will also hold for [itex]g_+[/itex] defined by

    1,\quad &f(x)-h(x)>0\\
    0,\quad &f(x)-h(x)\leq 0\\

    and also for [itex]g_-[/itex] defined similarly.
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