I want to prove the following: If $f$ is a bounded function on $[0,2\pi]$ and $A\subset[0,2\pi]$ a measurable set, then $\forall\epsilon>0$ $\exists$ a finite union $U$ of open intervals s.t. $|\int_Uf-\int_Af|<\epsilon$.
I have difficulties to show, that there is a $U$ which has such a form. I considered two cases:
- If $(0,\pi)\subset A\subset[0,\pi]$, take $U=(0,2\pi)$, then $\mu(A\setminus U)=0$ and I can show the claim. (I even don't need that $f$ is bounded for this case?)
- In the other case, I know that $\forall\delta>0\ \exists U\subset[0,2\pi]$ open s.t. $A\subset U$ and $\mu(U\setminus A)<\delta$ because $A$ is measurable. If I take $\delta=\epsilon/M$ (where $M$ is the bound of $f$) I can show the claim for this $U$. But in general this is an arbitrary union of open intervals? How can I find a finite one?
It seems to me that it would be easier to show this by regularity of the Lebesgue measure $\lambda$. You know that since $f$ is bounded, that there exists $\delta>0$ such that if $\lambda(E)<\delta$ then $\int_E \vert f\vert< \frac{1}{4} \epsilon$. By regularity there exists a closed subset $F\subseteq A$ such that $\lambda(A\setminus F)< \frac{\delta}{2} $. $F$ is a compact set, hence for the open cover $\{ (x-\frac{\delta}{2}),x+\frac{\delta}{2} \}_{x\in F} $, there exists a finite subcover of open intervals, and define it's union $U$. The measure of $U\Delta A$ will be less than $\delta$, hence by the triangle inequality you will obtain what you need.
Approximating $A$ by a closed subset (which is also compact) is the additional step you need.