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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:

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    $\begingroup$ You mentioned probability theory here, but it seems the measure you are considering is arbitrary, and this seems to make a difference! If for example the measure is finite, then $\mu_*(E^c) = \mu(\Omega) - \mu^*( E)$, which handles the complement case with a little manipulation. The general union case should follow from disjoint unions, since any union of sets can be written as a disjoint union ( $\cup_i A_i = \cup_i \cap^{i-1}_{j=1} A_j^c \cap A_i$). $\endgroup$ Sep 6 '21 at 20:56
  • $\begingroup$ You're right, finitely valued measures make a big difference and simplify. I still don't see how you show that $\mathcal{B}$ is closed over complement. With sub/super additivity of the outer/inner measures you get $\mu_*(E) + \mu_*(E^c) \le \mu(\Omega) \le \mu^*(E) + \mu^*(E^c)$. What you wrote, $\mu_*(E^c) = \mu(\Omega) - \mu^*(E)$ implies that the inner measure is finitely additive rather than merely super-additive which isn't a property that we have. $\endgroup$
    – clay
    Sep 6 '21 at 22:44
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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, $\mu^*(A_1 \cap A_2)=\mu_*(A_1\cap A_2)$.

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  • $\begingroup$ In the first paragraph you say $\cup_i B_i \subset A$, then later $(\cup_i B_i)^c \subset A^c$. Is that a mistake? $\endgroup$
    – clay
    Sep 7 '21 at 21:41
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    $\begingroup$ Yes that is a typo! Just fixed it. $\endgroup$ Sep 7 '21 at 22:17
  • $\begingroup$ ok, you've demonstrated that $\mathcal{B}$ is closed over complement. I've shown that it is closed over countable disjoint unions. We haven't demonstrated closure over general unions or intersections. If you demonstrate one of those it is trivial to derive the other. However, if you don't have either, it's not obvious how to do so. $\endgroup$
    – clay
    Sep 8 '21 at 1:46
  • $\begingroup$ I mentioned in my previous comment that the general union case follows from the disjoint union case, since every union of sets in $\mathcal{A}$ can be written as a disjoint union of sets in $\mathcal{A}$, constructed from a finite number of complements/intersections. So in essence you have already done that! And what you wrote looks good to me. $\endgroup$ Sep 8 '21 at 2:35
  • $\begingroup$ You did mention that. And you showed a formula that uses intersection. We haven't demonstrated closure over intersection. $\endgroup$
    – clay
    Sep 8 '21 at 2:47

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