On space $\Omega$ we have algebra $\mathcal{A} \subset \mathcal{P}(\Omega)$ with measure $\mu: \mathcal{A} \to [0,1]$ and we define the inner measure $\mu_*: \mathcal{P}(\Omega) \to [0,1]$ and outer measure $\mu^*: \mathcal{P}(\Omega) \to [0,1]$ as:
\begin{align*}
\mu_*(A) &= \text{sup} \left\{ \sum\limits_{i=1}^\infty \mu(B_i) : \text{for all disjoint sequences such that} \; B_i \in \mathcal{A}, \; \cup_{i = 1}^\infty B_i \subset A \right\} \\
\mu^*(A) &= \text{inf} \left\{ \sum\limits_{i=1}^\infty \mu(C_i) : \text{for all sequences such that} \; C_i \in \mathcal{A}, A \subset \cup_{i=1}^\infty C_i \right\} \\
\end{align*}Show that the collection of sets $\mathcal{B}$ where $\mu_*(A) = \mu^*(A)$ is a $\sigma$ algebra.
A few properties:
Both inner and outer measure are monotonic such that if $A \subset B$, then $\mu_*(A) \le \mu_*(B)$ and $\mu^*(A) \le \mu^*(B)$
For any $A \in \mathcal{A}$, $\mu_*(A) = \mu(A) = \mu^*(A)$ such that $\mathcal{A} \subset \mathcal{B}$.
For any $A \subset \Omega$, $\mu_*(A) \le \mu^*(A)$.
The inner measure is superadditive for disjoint $A_i \subset \Omega$ and the outer measure is subadditive for any $A_i \subset \Omega$ such that:
\begin{align*}
\mu_* \left( \cup_{i \ge 1} A_i \right) \ge \sum_{i \ge 1} \mu_*(A_i) \\
\mu^* \left( \cup_{i \ge 1} A_i \right) \le \sum_{i \ge 1} \mu^*(A_i) \\
\end{align*}
For any $A \in \Omega$, $\mu_*(A) \le \mu^*(A)$ so, where $A_i$ are disjoint, we have:
\begin{align*}
\sum_{i \ge 1} \mu_*(A_i) \le \mu_* \left( \cup_{i \ge 1} A_i \right) \le \mu^* \left( \cup_{i \ge 1} A_i \right) \le \sum_{i \ge 1} \mu^*(A_i) \\
\end{align*}
For any disjoint countable $A_i \in \mathcal{B}$ such that $\mu_*(A_i) = \mu^*(A_i)$, then $\sum_{i \ge 1} \mu_*(A_i) = \sum_{i \ge 1} \mu^*(A_i)$ such that:
\begin{align*}
\mu_* \left( \cup_{i \ge 1} A_i \right) &= \mu^* \left( \cup_{i \ge 1} A_i \right) \\
\end{align*}
Therefore, $\mathcal{B}$ is closed over countable disjoint union.
From the definitions of inner/outer measure, for any such sequences $B_i, C_i \in \mathcal{A} \subset \mathcal{B}$, we have:
\begin{align*}
\cup_{i = 1}^\infty B_i \subset A \subset \cup_{i=1}^\infty C_i \\
\end{align*}
With monotonicity of both the inner and outer measures:
\begin{align*}
\mu_*(\cup_{i = 1}^\infty B_i) &\le \mu_*(A) \le \mu_*(\cup_{i=1}^\infty C_i) \\
\mu^*(\cup_{i = 1}^\infty B_i) &\le \mu^*(A) \le \mu^*(\cup_{i=1}^\infty C_i) \\
\end{align*}
Combining with $\mu_*(A) \le \mu^*(A)$ and also $\mu_*(B_i) = \mu^*(B_i)$ and $\mu_*(C_i) = \mu^*(C_i)$
\begin{align*}
\mu_*(\cup_{i = 1}^\infty B_i) = \mu^*(\cup_{i = 1}^\infty B_i) \le \mu_*(A) \le \mu^*(A) \le \mu_*(\cup_{i=1}^\infty C_i) = \mu^*(\cup_{i=1}^\infty C_i) \\
\end{align*}
For any $E \in \mathcal{B}$ such that $\mu_*(E) = \mu^*(E)$ superadditivity and subadditivity give us:
\begin{align*}
\mu_*(E) + \mu_*(E^c) \le \mu(\Omega) \le \mu^*(E) + \mu^*(E^c) \\
\mu_*(E^c) \le \mu(\Omega) – \mu_*(E) \le \mu^*(E^c) \\
\end{align*}
From there, how do we demonstrate that $\mu_*(E^) = \mu^*(E^c)$?
Now, I'm stuck on what to do. I need to show that $\mathcal{B}$ is closed over complement and countable (not necessarily disjoint) union to say that it is a $\sigma$-measure.
BTW, with stackexchange search, I see this question asked twice before with no good solutions:
Best Answer
Now that the measures are assumed finite I think this is doable. First we wish to show that $\mu_*(A) = \mu(\Omega) - \mu^*(A)$. Let $\epsilon > 0$. Choose $B_i$ so that $\cup_i B_i \subset A$, and $\mu_*(A) \le \mu(\cup_i B_i) + \epsilon$. It follows since $\mu$ is additive that for $C\in \mathcal{A}$, $\mu(C) = \mu(\Omega) - \mu(C^c)$. A picky detail, but additionally there exists $n$ so that $\mu(\cup_i B_i) \le \mu(\cup_{i=1}^n B_i) + \epsilon$. This is needed since $\cup_{i=1}^n B_i \in \mathcal{A}$ ($\mathcal{A}$ is an algebra). With this, and noting that $ A^c \subset \left(\cup_{i=1}^n B_i \right)^c $, $$ \mu_*(A) \le \mu(\cup_i B_i) + \epsilon \le \mu(\cup_{i=1}^n B_i) + 2\epsilon = \mu(\Omega) - \mu(\left(\cup_{i=1}^n B_i \right)^c) + 2\epsilon \le \mu(\Omega) - \mu^*(A^c) + 2\epsilon. $$ Starting with a cover $C_i$ such that $A^c \subset \cup_i C_i$ and $\mu(\cup_i C_i) \le \mu^*(A^c) + \epsilon$, one can show very similarly that $$ \mu^*(A^c) \ge \mu(\Omega) - \mu_*(A) -2\epsilon. $$ Manipulating a bit and putting the above equations together we have $$ \mu(\Omega)-\mu^*(A^c) - 2\epsilon \le \mu_*(A) \le \mu(\Omega)-\mu^*(A^c) + 2\epsilon. $$ Since $\epsilon$ is arbitrary, we have $\mu_*(A) = \mu(\Omega)-\mu^*(A^c)$. Now this implies that the collection of sets $\mathcal{B}$ is closed under the complement. Suppose $A\in \mathcal{B}$ so that $\mu_*(A)= \mu^*(A)$. Then $\mu_*(A^c) =\mu(\Omega)-\mu^*(A)= \mu(\Omega)-\mu_*(A)= \mu^*(A^c)$.
It also came up in the comments to show that $\mathcal{B}$ is closed under intersection. To that end let $\epsilon>0$, and suppose $A_1,A_2 \in \mathcal{B}$. As a result we may choose $B_{i,1},B_{i,2},C_{i,2},C_{i,2}$ with the following properties: 1) $\cup_i B_{i,j} \subset A_j \subset \cup_i C_{i,j}$, $j=1,2$. 2) $\mu( \cup_{i} C_{i,j} / \cup_{i} B_{i,j}) < \epsilon/2$, $j=1,2$. Evidently then $(\cup_i B_{i,1}) \cap (\cup_i B_{i,2}) \subset A_1 \cap A_2 \subset (\cup_i C_{i,1}) \cap (\cup_i C_{i,2})$. Further, it can be shown that $(\cup_{i } C_{i,1} \cap C_{i,2}) /(\cup_{i } B_{i,1} \cap B_{i,2}) \subset (\cup_{i} C_{i,1} / \cup_{i} B_{i,1}) \cup (\cup_{i} C_{i,2} / \cup_{i} B_{i,2}). $ It follows that $\mu ( (\cup_i C_{i,1} \cap \cup_i C_{i,2}) /(\cup_i B_{i,1}) \cap (\cup_i B_{i,2})) < \epsilon$, giving that $\mu^*(A_1 \cap A_2) - \mu_*(A_1\cap A_2) < \epsilon$. Since $\epsilon$ is arbitrary, $\mu^*(A_1 \cap A_2)=\mu_*(A_1\cap A_2)$.