Show that there exist $G,H \in \sigma (\mathcal A_0)$ with $H \subseteq A \subseteq G$ such that $\mu^* (G \setminus H) = 0.$

Question

Let $\mathcal A_0$ be an algebra of subsets of a set $X$ and let $\mu$ be a finite measure on $\mathcal A_0.$ Let $\mu^*$ be the outer measure induced by $\mu.$ Let $A \subseteq X$ be such that for any given $\varepsilon \gt 0$ there exists $A_{\varepsilon} \in \mathcal A_0$ such that $\mu^*(A\ \Delta\ A_{\varepsilon}) \lt \varepsilon.$ Then show that there exist two sets $G,H \in \sigma (\mathcal A_0)$ with $H \subseteq A \subseteq G$ such that $\mu^*(G \setminus H) = 0.$

[Here $\sigma (\mathcal A_0)$ denotes the $\sigma$-algebra generated by $\mathcal A_0$].

Let $\widetilde {\mathcal A}$ be the collection of all countable unions of elements of $\mathcal A_0.$ Then for any $E \subseteq X$ we define $\mu^*(E)$ as follows $:$ $$\mu^* (E) : = \inf \left \{\overline {\mu} (G)\ \bigg |\ G \supseteq E, G \in \widetilde {\mathcal A} \right \}.$$ From this definition and from the given hypothesis how do I get hold of $G,H \in \mathcal \sigma (\mathcal A_0)\ $? Any help in this regard will be appreciated.

Thanks in advance.

Answer 1

Let us start by proving two lemmas.

Lemma 1: For all $E\subseteq X$, there is $G \in \sigma (\mathcal A_0)$ such that $E\subseteq G$ and $ \mu^*(E) = \overline {\mu} (G)$

Proof of Lemma 1: Note that if $E \subseteq F$ then $\mu^*(E) \leq \mu^*(F)$. So, for all $E\subseteq X$, $\mu^*(E) \leq \mu^*(X) = \mu(X) < \infty$.

By definition:
$$\mu^* (E) : = \inf \left \{\overline {\mu} (G)\ \bigg |\ G \supseteq E, G \in \widetilde {\mathcal A} \right \}$$ where $\overline {\mu} $ is the extension of $\mu$ to $\sigma (\mathcal A_0)$.

So, there is $\{G_n\}_n$ sequence in $ \widetilde {\mathcal A}$, such that, for all $n$ , $G_n \supseteq E$ and $\overline {\mu} (G_n) < \mu^* (E) + \frac{1}{n}$.

Now, let $G = \bigcap_n G_n$. Then, $E \subseteq G $ and, since $G \in \sigma (\mathcal A_0)$, we have that $\mu^*(G) = \overline {\mu} (G) $. So, we have $$ \mu^*(E) \leq \mu^*(G) = \overline {\mu} (G) \leq \overline {\mu} (G_n) < \mu^* (E) + \frac{1}{n}$$

for all $n$. So, $ \mu^*(E) = \mu^*(G) =\overline {\mu} (G)$.

Lemma 2: For all $A \subseteq X$, if, for any given $\varepsilon \gt 0$, there exists $A_{\varepsilon} \in \mathcal A_0$ such that $\mu^*(A\ \Delta\ A_{\varepsilon}) \lt \varepsilon$, then $A$ is $\mu^*$-measurable. $\square$

Proof of Lemma 2: Given any $\varepsilon \gt 0$, note that $$A \cup A_{\varepsilon} = A_{\varepsilon} \cup (A\ \Delta\ A_{\varepsilon}) \tag{1}$$ and $$ A^c \cup A_{\varepsilon}^c = A_{\varepsilon}^c \cup (A^c \setminus A_{\varepsilon}^c)= A_{\varepsilon}^c \cup ( A_{\varepsilon} \setminus A) \subseteq A_{\varepsilon}^c \cup (A\ \Delta\ A_{\varepsilon}) \tag{2}$$

Now, given any $C\subseteq X$, for all $\varepsilon \gt 0$, we have that $A_{\varepsilon} \in \mathcal A_0 \subseteq \sigma(\mathcal A_0)$, so $$\mu^*(C) = \mu^*(C \cap A_{\varepsilon}) + \mu^*(C \cap A_{\varepsilon}^c) \tag{3}$$

Using $(1)$, $(2)$ and $(3)$:

\begin{align*} \mu^*(C) & \leq \mu^*(C \cap A )+ \mu^*(C \cap A^c ) \leq \\ &\leq \mu^*(C \cap (A \cup A_{\varepsilon})) + \mu^*(C \cap (A^c \cup A_{\varepsilon}^c)) \leq \\ & \leq \mu^*(C \cap (A_{\varepsilon} \cup (A\ \Delta\ A_{\varepsilon}))) + \mu^*(C \cap (A_{\varepsilon}^c \cup (A\ \Delta\ A_{\varepsilon}))) \leq \\ & \leq \mu^*(C \cap A_{\varepsilon}) + \mu^*(C \cap (A\ \Delta\ A_{\varepsilon})) + \mu^*(C \cap A_{\varepsilon}^c) + \mu^*(C \cap (A\ \Delta\ A_{\varepsilon})) = \\ & = \mu^*(C) + 2 \mu^*(C \cap (A\ \Delta\ A_{\varepsilon})) < \\ &< \mu^*(C) + 2 \varepsilon \end{align*} Since this is true for all $\varepsilon \gt 0$, we have that $$ \mu^*(C) = \mu^*(C \cap A )+ \mu^*(C \cap A^c ) $$ Since this is true for all $C\subseteq X$, we have that $A$ is $\mu^*$-measurable. $\square$

Now let us prove the main result:

Let $\mathcal A_0$ be an algebra of subsets of a set $X$ and let $\mu$ be a finite measure on $\mathcal A_0.$ Let $\mu^*$ be the outer measure induced by $\mu.$ Let $A \subseteq X$ be such that for any given $\varepsilon \gt 0$ there exists $A_{\varepsilon} \in \mathcal A_0$ such that $\mu^*(A\ \Delta\ A_{\varepsilon}) \lt \varepsilon.$ Then show that there exist two sets $G,H \in \sigma (\mathcal A_0)$ with $H \subseteq A \subseteq G$ such that $\mu^*(G \setminus H) = 0.$

[Here $\sigma (\mathcal A_0)$ denotes the $\sigma$-algebra generated by $\mathcal A_0$].

Proof: From lemma 1, there are $G, F \in \sigma (\mathcal A_0)$ such that:

1. $A\subseteq G$ and $ \mu^*(A) = \overline {\mu} (G)$
2. $A^c\subseteq F$ and $ \mu^*(A^c) = \overline {\mu} (F)$

Note that $$ X = A \cup A^c \subseteq G \cup F \subseteq X$$ So $X= G \cup F$.

From lemma 2, taking $C=X$, we have $$ \mu^*(X) = \mu^*(A )+ \mu^*( A^c ) $$

So we have $$ \overline {\mu} (G \cup F) = \mu^*(G \cup F)= \mu^*(X) = \mu^*(A )+ \mu^*( A^c )= \overline {\mu}(G) + \overline {\mu}(F)$$

So $\overline {\mu}(G \cap F) = 0$. Taking $H=F^c$, we have that there are $G, H \in \sigma (\mathcal A_0)$, $ H \subseteq A \subseteq G$ and $\mu^*(G \setminus H)=\overline {\mu}(G \setminus H) =\overline {\mu}(G \cap F) = 0$. $\square$