I have the following question from an elementary course on Measure Theory:
Let $E$ be a measurable set with $m(E)<\infty$. Show that there is a descending sequence of open sets $\{G_n\}$ so that $E\subseteq G_n$ for all $n \ \in \mathbb N$ and $ \lim_{n\to\infty} m(G_n)=m(E)$.
What I've done so far:
We know that there is a $G_{\delta}$ - set $G_1$ such that $E\subseteq G_1$ and $m(G_1\setminus E)=0$. Then since $G_1$ is measurable we can find a set $G_2$ satisfying the same conditions. Continuing this way we can find an ascending sequence of sets. I don't know how to get a descending one. Am I on the right track or should I try a different approach?
Any hints/ideas are much appreciated.
Thanks in advance for any replies.
Since E is measurable, for each $ n\in \mathbb{N} $ , there is an open set $E_{n}$ such that $ E\subseteq E_{n} $ and $m(E_n\setminus E)<1/n$. Define $ G_{n}=\cap_{j=1}^{n}E_{j} $ for each $ n\in \mathbb{N} $. Then for each $ n\in \mathbb{N} $, $ G_{n}$ is open and {$ G_{n}$} is a descending sequence. Since $ G_{n}\subseteq E_{n} $ , we have $m(G_n\setminus E)<1/n$ for each $n\in \mathbb{N}$. Since $m(E)<\infty$ by excision property of measure also we have $m(G_n)-m(E)<1/n.$
So we have that $$ \lim_{n\to\infty} m(G_n)=m(E)$$ $\square$