Sets such that inner measure equals outer measure is a sigma algebra Let $\mu$ be a measure on a $\sigma$-algebra $\mathcal{A}$ in $\mathcal{P}(\Omega)$ such that $\mu(\Omega)&lt;\infty$. Define inner and outer measures as follows.

$\mu^*(E)=\inf\{\sum_{n=1}^{\infty}\mu(A_n): A_n\in\mathcal{A}, E\subset\bigcup_{n=1}^{\infty}A_n\}$

and

$\mu_*(E)=\sup\{\mu(A): A\in\mathcal{A}, A\subset E\}$.
I need to prove that the set $\bar{\mathcal{A}}=\{E\subset\Omega: \mu_*(E)=\mu^*(E)\}$ is a $\sigma$-algebra. 

First, I tried to prove it in a direct manner. But I got stuck. 

After trying some other exercises I thought, $\bar{\mathcal{A}}=\{E\subset\Omega: \mu_*(E)=\mu^*(E)\}=\{E\subset\Omega: \exists A,B\in\mathcal{A}, A\subset E\subset B, \mu(B\setminus A)=0\}$. Is this true? 

Should I prove the first one direct, or can I prove that the second one is a $\sigma$-algebra?
If $A_1,A_2,\dots$ is a sequence in $\mathcal A$ with $E\subseteq\bigcup_{n=1}^{\infty}A_n$ and $B_n=A_n-\bigcup_{i=1}^{n-1}A_i$ then $B_1,B_2,\dots$ is a sequence in $\mathcal A$ with $E\subseteq\bigcup_{n=1}^{\infty}A_n=\bigcup_{n=1}^{\infty}B_n$.

Setting $B=\bigcup_{n=1}^{\infty}B_n$ we find that $B\in\mathcal A$ with $$\mu^*(E)\leq\mu(B)=\sum_{n=1}^{\infty}\mu(B_n)\leq\sum_{n=1}^{\infty}\mu(A_n)$$

Then $\mu^*(E)&lt;\infty$ and this also makes clear that for every integer $k$ we can find a set $C_k\in\mathcal A$ with: $$\mu^*(E)\leq\mu(C_k)\leq\mu^*(E)+\frac1{k}$$

Then $C:=\bigcap_{k=1}^{\infty}C_k\in\mathcal A$ with $E\subseteq C$ and $\mu^*(E)=\mu(C)$.

Also we can find a sequence $D_1,D_2,\dots$ in $\mathcal A$ with $D_1\subseteq D_2\subseteq\cdots\subseteq E$ and $\mu(D_n)\to\mu_*(E)$.

Then $D:=\bigcup_{n=1}^{\infty}D_n\in\mathcal A$ with $D\subseteq E\subseteq C$ and $\mu(D)=\mu_*(E)$.

Then also $\mu(C-D)=0$.

Proved is now that $\bar{\mathcal{A}}\subseteq\{E\subset\Omega: \exists C,D\in\mathcal{A}, D\subseteq E\subseteq C, \mu(C-D)=0\}$.

The side $\supseteq$ is quite evident.
If $A_1,A_2,\dots$ is a sequence in $\mathcal A$ with $E\subseteq\bigcup_{n=1}^{\infty}A_n$ and $B_n=A_n-\bigcup_{i=1}^{n-1}A_i$ then $B_1,B_2,\dots$ is a sequence in $\mathcal A$ with $E\subseteq\bigcup_{n=1}^{\infty}A_n=\bigcup_{n=1}^{\infty}B_n$.

Setting $B=\bigcup_{n=1}^{\infty}B_n$ we find that $B\in\mathcal A$ with $$\mu^*(E)\leq\mu(B)=\sum_{n=1}^{\infty}\mu(B_n)\leq\sum_{n=1}^{\infty}\mu(A_n)$$

Then $\mu^*(E)&lt;\infty$ and this also makes clear that for every integer $k$ we can find a set $C_k\in\mathcal A$ with: $$\mu^*(E)\leq\mu(C_k)\leq\mu^*(E)+\frac1{k}$$

Then $C:=\bigcap_{k=1}^{\infty}C_k\in\mathcal A$ with $E\subseteq C$ and $\mu^*(E)=\mu(C)$.

Also we can find a sequence $D_1,D_2,\dots$ in $\mathcal A$ with $D_1\subseteq D_2\subseteq\cdots\subseteq E$ and $\mu(D_n)\to\mu_*(E)$.

Then $D:=\bigcup_{n=1}^{\infty}D_n\in\mathcal A$ with $D\subseteq E\subseteq C$ and $\mu(D)=\mu_*(E)$.

Then also $\mu(C-D)=0$.

Proved is now that $\bar{\mathcal{A}}\subseteq\{E\subset\Omega: \exists C,D\in\mathcal{A}, D\subseteq E\subseteq C, \mu(C-D)=0\}$.

The side $\supseteq$ is quite evident.