If three measurable sets $A,B,C$ have the relation of $A\subseteq B\subseteq C$,$B$ is closed and $C$ is compact, with $m(A)=m(C)=1$, how can we know that $m(B)=1$?
I've found that the properties of outer measures says if $A\subseteq B$, then $m^{*}(A)\leq m^{*}(B)$, but can we just use $m^{*}(B)$ to get $m(B)$ as we don't know whether $B$ is bounded or not, so $m^{*}(B)$ is not necessarily the same as $m(B)$.
By the measure function properties, we know that given $A,B$ measurable sets, $$A\subseteq B \implies m(A)\leq m(B).$$
So, if $A\subseteq B\subseteq C$, then $m(A)\leq m(B)\leq m(C)$.
If $m(A)=m(C)=1$, then $1\leq m(B)\leq 1$, and by Sandwich's Theorem we know that $$m(B)=1.$$