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hello im having trouble showing the following:

let $u$ be a positive measure. if $\int_E f\, du= \int_E g\, du$ for all measurable $E$ then $f=g$ a.e.

i was trying to argue by contradiction: if $f\neq g$ a.e. then there must exist some set $E=\{x: f(x)\neq g(x)\}$ such that $u(E) \gt 0$. Then let $E^+=\{x: f(x)\gt g(x)\}$ and $E^-=\{x: f(x)\lt g(x)\}$. Now, if $E^+$ or $E^-$ is measurable and have positive measure then $\int_{E^+} f\, du \gt \int_{E^+} g\, du$ or $\int_{E^-} f\, du \lt \int_{E^-} g\, du$, contradiction.

As you can see, the argument hinges on $E^+$ or $E^-$ being measurable. This is the part im having trouble with.

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The assertion that $f\ne g$ a.e. is not the negation of the assertion that $f=g$ a.e. –  Did Mar 8 '11 at 21:21
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2 Answers 2

up vote 2 down vote accepted

Hint We have that $f$ and $g$ map into $\mathbb{R}$ from some unknown measure space, say $(X,\mathcal{M})$. Let $h(x)=f(x)-g(x)$. Then $E^+\subset h^{-1}(0,\infty)$ and $E^- \subset h^{-1}(-\infty,0)$. Then recall that the sum of two measurable functions is measurable (addition is continuous), and $h$ being $\mathcal{M}$-measurable is equivalent to $f^{-1}(a,\infty)\in\mathcal{M}$ for every $a\in \mathbb{R}$ and it is also equivalent to $f^{-1}(-\infty, a)\in\mathcal{M}$ for every $a\in \mathbb{R}$. (Since these sets generate the Borel sigma algebra over $\mathbb{R}$)

Then conclude $h^{-1}(0,\infty)$ and $h^{-1}(-\infty,0)$ are measurable, and integrate over them.

Hope that helps,

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$E^+=h^{-1}((0,\infty))$ and $E^-=h^{-1}((-\infty,0))$. –  Did Mar 8 '11 at 21:20
    
@Didier Piau: Hmmm. I interpreted "there must exists some set $E=\{x: f(x)\neq g(x)\}$ such that $\mu (E)>0$" to mean "take some set $E$ with $\mu(E)\neq0$ where $f(x)\neq g(x)$ on $E$." In that case there exists many, many such $E$, but it is guaranteed that $E^+\subset h^{-1}(0,\infty)$. However the next line after that seems to be what you assert, "let $E^+ =\{x:f(x)>g(x)\}$..." –  Eric Naslund Mar 9 '11 at 3:02
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The OP defines the set $E$ itself as $E=\{x;f(x)\ne g(x)\}$, that is, $E=h^{-1}((-\infty,0)\cup(0,+\infty))$. The interpretation you mention runs into the problem that any subset of these $E$, $E^+$ and $E^-$ should be measurable (which is ludicrous). –  Did Mar 9 '11 at 6:35
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hint:

  1. The difference of two measurable functions is measurable
  2. $(0,\infty)\subset\mathbb{R}$ is Borel, so for a measurable function $F$, the set on which it takes positive values is a measurable set.
  3. The collection of measurable sets form a $\sigma$ algebra, and in particular intersection of two measurable sets is measurable.
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I'd hate to assume that this is a measure on $\mathbb{R}$ and defined on the Borel sets of the real lines. Good hints, though. –  Asaf Karagila Mar 8 '11 at 17:15
    
@Asaf: When we do integration, we assume the functions map into $\mathbb{R}$. (or $\mathbb{C}$) It can be from an arbitrary measure space, but we have to go into $\mathbb{R}$ or $\mathbb{C}$. –  Eric Naslund Mar 8 '11 at 17:36
    
@Eric: When I took the course in measure theory, when we said a measurable function we only talked about a function between two measure spaces that the preimage of a measurable set is a measurable set. Not necessarily real or complex. –  Asaf Karagila Mar 8 '11 at 17:38
    
@Asaf: ... and then surely you studied the Lebesgue integral where the function takes value in some finite dimensional vector space, which can be projected down to $\mathbb{R}$. Furthermore from the steps outlined by the OP it was pretty clear that the functions take values in $\mathbb{R}$ anyway. (Without a linear structure I don't see how you can define an integral.) –  Willie Wong Mar 8 '11 at 17:43
    
@Asaf: Yes measurable functions can be from arbitrary spaces, but we cannot integrate arbitrary measurable functions. You don't necessarily have simply functions that can well approximate other functions (you don't necessarily have addition....) which is how the integral was defined. (Supremum over simple functions) –  Eric Naslund Mar 8 '11 at 17:58
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