Suppose $\mu^*$ is an outer measure induced from a premeasure on a ring generated by the semiring $\mathcal{H}$. Then one can show that $\mu^*$ is regular. How to prove that $\mu^*$ is continuous from below, i.e, if $E_n \subset E_{n+1}$ and $E_n \to E$, then $\mu^*(E_n) \to \mu^*(E)$?
Similar question is asked here. But I could not understand a certain part of the accepted answer. A user claimed that
$$ \mu^*(F_n\setminus E_n)+\mu^*(F_{n-1}\setminus E_{n-1})=0$$
where $F_n$ is a $\mu^*$-measurable set containing $E_n$ and satisfying $\mu^*(E_n)=\mu^*(F_n)$.
In other words, it is claimed that $\mu^*(F_n \setminus E_n)=\mu^*(F_{n-1} \setminus E_{n-1})=0$
How is this so?
One possible solution is $\mu^*(F_n \setminus E_n)=\mu^*(F_n) \setminus \mu^*(E_n)=0$. But this is not necessarily true for outer measures, and even in our case after restricting the domain. If $F_n \setminus E_n$ is $\mu^*$-measurable, the claim is true, but this set difference is not $\mu^*$-measurable.
Idea?