3
$\begingroup$

Hello I'm 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 I'm having trouble with.

  • 0
    The assertion that $f\ne g$ a.e. is not the negation of the assertion that $f=g$ a.e.2019-02-04

2 Answers 2

2

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,

  • 1
    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).2011-03-09
4

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.
  • 0
    Willie: The parenthesis causes a compilation error :-) Also, I thought about it, and my remark is indeed erroneous.2011-03-08