Let $(X,\mathfrak{M},\mu)$ be a measure space and $\mathfrak{M}^*=\{E\subset X|\exists A,B\in\mathfrak{M} \text{ such that} A\subset E \subset B \text{ and} \mu(B\setminus A)=0\}$.
How do i show that $\mathfrak{M}^*$ is closed under countable union?
Following is what i tried:
Let $\{E_n\}$ be a sequence of elements of $\mathfrak{M}^*$.
Define $I_n=\{(A,B)\in\mathfrak{M}\times\mathfrak{M}|A\subset E_n\subset B \text{ and} \mu(B\setminus A)=0\}$
Again, define $P_n=\{A\in \mathfrak{M}|(A,B)\in I_n\}$ and $Q_n=\{B\in \mathfrak{M}|(A,B)\in I_n\}$.
Then, $\bigcup P_n$ and $\bigcap Q_n$ are elements of $\mathfrak{M}^*$ for every $n\in\mathbb{N}$.
It can be shown that [If $C$ is measurable and $\bigcup P_n\subset C \subset E_n$, then $C\subset \bigcup P_n$], which means that there is no measurable set between $\bigcup P_n$ and $E_n$.
Is this $P_n$ and $Q_n$ are in $\mathfrak{M}$?