I am stuck with proving the following theorem.
$A \subseteq \mathbb{R}$ is Lebesgue measurable iff for every $\varepsilon > 0$, there exist an open set $U \subseteq \mathbb{R}$ and a closed set $F \subseteq \mathbb{R}$ with $F \subseteq A \subseteq U$ such that $m(U \setminus F) < \varepsilon$.
I have managed to prove the case when $A$ is bounded. But I have struggled to prove the case when $A$ is unbounded. Here is how far I managed to get.
Assume $A$ is Lebesgue measurable and unbounded, then as I proved the case for bounded $A$, for every $\varepsilon > 0$, there exist an open set $U_n \subseteq \mathbb{R}$ and an closed set $F_n \subseteq \mathbb{R}$ with $F_n \subseteq A \cap [-n,n] \subseteq U_n$ such that $m(U_n \setminus F_n) < \frac{\varepsilon}{2^n}$ for every natural number $n$. So I was told to let $U = \bigcup_{n=1}^{\infty}U_n$ and $F = \bigcup_{n=1}^{\infty}(F_n \cap ([-n,-n+1] \cup [n-1,n]))$.
It is easy to see $U$ is open as $U$ is the union of open sets. But I struggle to see why $F$ is closed. Here is my attempt.
$[-n,-n+1] \cup [n-1,n]$ is closed because a finite union of closed sets is closed, then $F_n \cap ([-n,-n+1] \cup [n-1,n])$ is also closed since the intersection of closed sets is closed. But then why is $\bigcup_{n=1}^{\infty}(F_n \cap ([-n,-n+1] \cup [n-1,n]))$ is also closed?
Any help will be greatly appreciated.
Best Answer
Set $R_n := F_n\cap ([-n,-n+1]\cup [n-1,n])$ and let $(x_k)$ be a sequence in $F$ (the union) which converges to some $x$. We have to show that $x\in F$. Assume, e.g., that $x\in (n-1,n)$ for some $n > 0$. Then also $x_k\in (n-1,n)$ for $k\ge K$, so $x_k\in R_n$ for $k\ge K$ (it must be in some $R_\ell$ and only $R_n$ is possible). Hence, $x\in R_n\subset F$ since $R_n$ is closed. You can do the same for $x\in (-n,-n+1)$, of course.
So, you still have to figure out how to deal with the case that $x = n$ is an integer. But you can use the same principle here with the exception that in the end $x\in R_{n-1}\cup R_n\cup R_{n+1}$ is possible.