Can we say that $\mathcal A = \mathcal A^*\ $?

Question

Let $\mathcal A_0$ be an algebra of subsets of a set $X$ and let $\mu$ be a probability measure on $\mathcal A.$ Let $\mu^*$ be the outer measure induced by $\mu$ such that for all $A \subseteq X$ we have $$\mu^*(A) : = \inf \left \{\mu^*(G)\ \bigg |\ G \supseteq A, G \in \widetilde {\mathcal A} \right \}$$ where $\widetilde {\mathcal A}$ is the collection of countable unions of elements of $\mathcal A_0.$ Consider the collection $\mathcal A : = \left \{A \subseteq X\ \bigg |\ \mu^*(A) + \mu^*(A^c) = 1 \right \}.$ Then we can prove that $\mathcal A$ is a $\sigma$-algebra of subsets of $X$ containing $\mathcal A_0$ and $\mu^*$ is a measure on $\mathcal A.$ Let $\sigma (\mathcal A_0)$ denote the $\sigma$-algebra generated by $\mathcal A_0.$ Then clearly $\sigma (\mathcal A_0) \subseteq \mathcal A.$ The elements of $\mathcal A$ are called Lebesgue $\mu$-measurable subsets of $X.$

Now consider another collection $\mathcal A^*$ defined by $$\mathcal A^* : = \left \{A \subseteq X\ \bigg |\ \mu^* (A \cap E) + \mu^* (A^c \cap E) = \mu^* (E),\ \text {for all}\ E \subseteq X \right \}.$$ Then $\mathcal A^*$ can also be proved to be a $\sigma$-algebra of subsets of $X$ containing $\mathcal A_0.$

Question $:$ What is the relation between $\mathcal A$ and $\mathcal A^*\ $? Replacing $E$ by $X$ in the definition of $\mathcal A^*$ we find that $\mathcal A^* \subseteq \mathcal A.$ Does the reverse inclusion also hold in general? Can anybody come up with a counter-example?

Thanks in advance.

Answer 1

We can actually show that $\mathcal A = \mathcal A^*.$ As it is already shown that $\mathcal A^* \subseteq \mathcal A,$ so in order to prove the claim it is enough to show that $\mathcal A \subseteq \mathcal A^*.$

We will first show that $\sigma (\mathcal A_0) \subseteq \mathcal A^*.$ Since $\mathcal A^*$ is a $\sigma$-algebra, so it is enough to show that $\mathcal A_0 \subseteq \mathcal A^*.$ Let us take $A \in \mathcal A_0$ and $E \subseteq X.$ Let us take $\varepsilon \gt 0.$ Then by the definition of $\mu^*$ there exists $\widetilde {\mathcal A} \ni F \supseteq E$ such that $\mu^*(F) \lt \mu^*(E) + \varepsilon.$ Now $A, A^c, F \in \widetilde {\mathcal A} \subseteq \sigma (\mathcal A_0)$ and $\mu \rvert_{\mathcal \sigma (A_0)}$ is a measure, so by exploiting the finite additivity of $\mu^*$ on $\sigma (\mathcal A_0)$ we find that $$\mu^* (A \cap E) + \mu^* (A^c \cap E) \leq \mu^*(A \cap F) + \mu^* (A^c \cap F) = \mu^* (F) \lt \mu^* (E) + \varepsilon.$$ Since $\varepsilon \gt 0$ is arbitrarily taken it follows that $\mu^* (A \cap E) + \mu^* (A^c \cap E) \leq \mu^*(E).$ Also by countable subadditivity of $\mu^*$ it follows that $\mu^* (A \cap E) + \mu^* (A^c \cap E) \geq \mu^*(E).$ Combining these two inequalities it follows that $A \in \mathcal A^*$ and hence $\mathcal A_0 \subseteq \mathcal A^*,$ as required.

Now let $B \in \mathcal A.$ Then by the definition of $\mu^*$ we can get hold of $\widetilde {\mathcal A} \ni G \supseteq B$ such that $\mu^* (G) \lt \mu^* (B) + \varepsilon.$ Since $A, G \in \mathcal A$ and $\mu^* \rvert_\mathcal A$ is a measure we find that (by finite additivity of $\mu^*$ on $\mathcal A$) $$\mu^* (G \setminus B) = \mu^*(G) - \mu^*(B) \lt \varepsilon\ \ \ \ \ \ \ \ \ (1)$$ Then for any $E \subseteq X$ we have $$\begin{align*} \mu^* (E) & \leq \mu^* (B \cap E) + \mu^* (B^c \cap E) \\ & \leq \mu^* (G \cap E) + \mu^* ((G^c \cap E) \cup (G \cap B^c \cap E)) \\ & \leq \mu^*(G \cap E) + \mu^* (G^c \cap E) + \mu^* (G \cap B^c \cap E)\ \ (\because \mu^*\ \text {is countably subadditive}\ ) \\ & \leq \mu^* (E) + \mu^* (G \setminus B)\ \ (\because G \in \widetilde {\mathcal A} \subseteq \mathcal \sigma (\mathcal A_0) \subseteq \mathcal A^*\ \text {and}\ G \cap B^c \cap E \subseteq G \cap B^c = G \setminus B) \\ & \lt \mu^* (E) + \varepsilon\ \ (\text {by}\ (1)) \end{align*}$$ Since $\varepsilon \gt 0$ was arbitrarily taken it follows that $\mu^* (B \cap E) + \mu^*(B^c \cap E) = \mu^* (E).$ This shows that $B \in \mathcal A^*.$ Hence we have $\mathcal A \subseteq \mathcal A^*,$ as claimed.

QED