Vtyjkyuk

I want to prove the following: If $f$ is a bounded function on $[0,2pi]$ and $Asubset[0,2pi]$ a measurable set, then $forallepsilon>0$ $exists$ 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:

1. If $(0,pi)subset Asubset[0,pi]$, take $U=(0,2pi)$, then $mu(Asetminus U)=0$ and I can show the claim. (I even don't need that $f$ is bounded for this case?)


2. In the other case, I know that $foralldelta>0 exists Usubset[0,2pi]$ open s.t. $Asubset U$ and $mu(Usetminus A)<delta$ 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 $int_E vert fvert< frac14 epsilon$. By regularity there exists a closed subset $Fsubseteq A$ such that $lambda(Asetminus F)< fracdelta2 $. $F$ is a compact set, hence for the open cover $ (x-fracdelta2),x+fracdelta2 _xin F $, there exists a finite subcover of open intervals, and define it's union $U$. The measure of $UDelta A$ 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.