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.