$m^*(E)=q>0$, for any $c\in (0,q)$, there exist $E_0\subset E$, such that $m^*(E_0)=c$
$$m^*(E)=\inf\{mG|E\subset G ,\text{G is open set}\}$$
I think if I can find a set $A\subset E$, and $m^*(E-A)=c$, then I proof the question. But I can’t construct $A$.
can some help me
Let $f(x)=m^{*}(E \cap (-\infty,x])$. If $x <y$ then $(-\infty,y] \subset (-\infty,x] \cup (x,y]$. By monotnicity and subadditivity this gives $f(x) \leq f(y)\leq f(x) +(y-x)$. Hence $f$ is a continuous function. If we know that $f(x) \to 0$ as $x \to -\infty$ we can use IVP to finish the proof. For the general case note that $m^{*}(E \cap (-N,N)) \to q$ as $N \to \infty$. We can choose $N$ such that $m^{*}(E \cap (-N,N)) \in (c,q)$. Now apply the first case with $E \cap (-N,N)$ in place of $E$. [In this case $f(x)=0$ for all $x <-N$].