(a) Let $E\subset \mathbb{R}$. Show that exists $H\in G_{\delta}$ such that $E\subseteq H$ and $m^*(E)=m(H)$.
(b) Let $E\subset \mathbb{R}$ not necesary measurable with $m^*(E)<\infty$. Show that exists $H\in G_{\delta}$ with $E\subset H$ such that for any $M$ measurable set, $m^*(E\cap M)=m(H\cap M)$ holds.
My solution:
(a) By definition of infimum, for all $\epsilon>0$ exists $\bigcup_{n=1}^{\infty} I_n\supseteq E$ ($I_n$ open intervals) such that $\sum_{n=1}^{\infty} l(I_n)\leq m^*(E)+\epsilon$.
Let $G:=\bigcup_{n=1}^{\infty}I_n$ open then $m^*(G)\leq \sum_{n=1}^{\infty} l(I_n)\leq m^*(E)+\epsilon$
Therefore, for each $n$ natural, exists $G_n$ open set with $m^*(G_n)\leq m^*(E)+1/n$ then $m^*(G_n)\leq m^*(E)$.
Therefore, $H:=\bigcap_{n=1}^{\infty} G_n$ holds $m^*(H)=m^*(E)$ and $H$ is measurable then $m^*(H)=m(H)$.
(b) Here I am stuck. The idea is to demonstrate the following:
For (a), exitst $H$ such that $E\subset H$ and $m^*(E)=m^*(H)$ then $m^*(H\setminus E)=0$ then $m^*((H\cap M)\setminus (E\cap M))=0$ then $m^*(H\cap M)=m^*(E\cap M)$.
Therefore, if the property of excision were fulfilled for the outer measure, the result would be had but I don't know if it is valid in the outer outer measure.
Conjeture: If $A$, $B$ measurable and $A\subset B$ and $m^*(A)<\infty$ then $m^*(B-A)=m^*(B)-m^*(A)$.
Proof. $B=B\setminus A\cup A$ then $m^*(B)\leq m^*(B\setminus A)+m^*(A)$ then $m^*(B)-m^*(A)\leq m^*(B\setminus A)$ but i can't proves that $m^*(B)-m^*(A)\geq m^*(B\setminus A)$
How proves (b)?
Best Answer
Take the former $H$. Since $M$ is measurable, by Caratheodory condition we have $$m^*(E)=m^*(E\cap M)+m^*(E-M)\leq m^*(E\cap M)+m^*(H-M)$$ $$m^*(E\cap M)+m^*(H-M)\leq m^*(H\cap M)+m^*(H-M)=m^*(H)=m^*(E)$$ Therefore all the inequality can be changed into equality, and there you have it.
You conjecture is just the equivalent form of Caratheodory condition.