Im trying to disprove the principle for $\epsilon = 0$ so need to get measure of $(E\setminus G) \cup (G\setminus E)$ equal to $0$. Any suggestions for the counterexample?
Asked
Active
Viewed 118 times
0
-
1Littlewood's first principle, and a proof. – Julien Apr 14 '13 at 20:43
1 Answers
1
So, you want a set $E\subset\mathbb R$ of finite measure for which there is no open set $O$ which is the finite union of disjoint open intervals such that $m(O\triangle E)=0$.
Let $E$ be a "Fat" Cantor Set. If $O$ is nonempty, $O\setminus E$ contains an interval. If $O$ is empty, $E\setminus O$ has positive measure.
