Which of the following is/are correct way to find Lebesgue outer measure( measure) $m^*(A)$ for a Lebesgue measurable set $A\subseteq \Bbb R^n$?
$1.$ $m^*(A)=sup\{ m^*(B): B \subseteq A,\text{$B$ is open}\}$
$2.$ $m^*(A)=sup\{ m^*(B): B \subseteq A,\text{$B$ is closed}\}$
$3.$ $m^*(A)=sup\{ m^*(B): B \subseteq A,\text{$B$ is compact }\}$
$4.$ $m^*(A)=sup\{ m^*(B): B \subseteq A,\text{$B$ is Lebesgue measurable }\}$
$5.$ $m^*(A)=sup\{ m^*(B): B \subseteq A,\text{$B$ is Jordan measurable}\}$
I tried it as
Because $A$ is Lebesgue measurable so it’s inner and outer measure is same and hence option $2$ is correct. Option $4$ is trivial one . If I take Cantor like set $C$ then it has not open and Jordan measurable set of positive measure so option $1$ and $5$ does not work( Please correct me if I am wrong). I have no idea for option $3$ but it’s seems to be correct . Please help me to solve this problem. Thanks .
Best Answer
Your answers for 1), 2),4)and 5) are correct. For 3) just note that you can approximate $m^{*}(A)$ by a closed set $C$ contained in $A$ and consider the compact set $K=C\cap[ -N,N]$ with large enough $N$. I hope you can write out a proof using this hint.