\(\mu_0(\varnothing)=0\),
if \((A_n)_{n\in \mathbb N}\) is a sequence of disjoint sets in \(\mathcal A\) such that \(\bigcup_{n=1}^{\infty}A_n \in \mathcal A\), then \[\mu_0\bigg(\bigcup_{n=1}^{\infty}A_n\bigg)=\sum_{n=1}^{\infty}\mu_0(A_n).\]
Outer measures induced by premeasures.
If \(\mu_0\) is a premeasure on \(\mathcal A \subseteq \mathcal{P}(X)\), it induces an outer measure on \(X\): \[\mu^{*}(A)=\inf\bigg\{\sum_{j=1}^{\infty}\mu_0(E_j): (E_j)_{j\in\mathbb N} \subseteq \mathcal A \text{ and } A \subseteq \bigcup_{j=1}^{\infty}E_j\bigg\}, \qquad A\subseteq \mathcal P(X).\]
\(\mu_0(E)=\mu^{*}(E)\) for all \(E \in \mathcal A\),
\(\mathcal A\subseteq \mathcal M(\mu^*)\). In other words, every set in \(\mathcal A\) is \(\mu^{*}\)-measurable.
Proof of (a). Suppose \(E \in \mathcal A\), we will show that \(\mu_0(E)=\mu^{*}(E)\).
Taking \(E_1=E\) and \(E_2=E_3=\ldots=\varnothing\) we see that \(E \subseteq \bigcup_{j=1}^{\infty}E_j\) and consequently \(\mu^{*}(E) \leq \mu_0(E)\).
We now reverse this inequality and show that \(\mu_0(E) \leq \mu^{*}(E)\).
If \(E \subseteq \bigcup_{j=1}^{\infty}A_j\) with \(A_j \in \mathcal A\), then we let \[B_n=E \cap \bigg(A_n \setminus \bigcup_{j=1}^{n-1}A_j\bigg)\in\mathcal A.\] Then \(B_n \cap B_m =\ \varnothing\) if \(n \neq m\) and \(E\subseteq \bigcup_{j=1}^{\infty}B_j=\bigcup_{j=1}^{\infty}A_j\). Now \[\begin{aligned} \mu_0(E)\le \sum_{j=1}^{\infty}\mu_0(B_j) \leq \sum_{j=1}^{\infty}\mu_0(A_j). \end{aligned}\] It follows that \(\mu_0(E) \leq \mu^{*}(E)\) as claimed.
Proof of (b). We now prove that \(\mathcal A\subseteq \mathcal M(\mu^*)\).
Let \(C \in \mathcal A\) and \(E \subseteq X\) it suffices to show that \[\mu^{*}(E) \geq \mu^{*}(E \cap C)+\mu^{*}(E \cap C^c).\] Let \(\varepsilon>0\), then there is a sequence \((B_n)_{n=1}^{\infty}\subseteq \mathcal A\) such that \[E \subseteq \bigcup_{n =1}^{\infty}B_n \quad \text{ and } \quad \sum_{n=1}^{\infty}\mu_0(B_n) \leq \mu^{*}(E)+\varepsilon.\] \(\mu_0\) is additive on \(\mathcal A\), thus \[\begin{aligned} \mu^{*}(E)+\varepsilon \geq \sum_{j=1}^{\infty}\mu_0(B_j)= \sum_{j=1}^{\infty}(\mu_0(B_j \cap C)+\mu_0(B_j \cap C^c)). \end{aligned}\]
Since \(\mu_0 \geq \mu^{*}\) and \(B=\bigcup_{j=1}^{\infty}B_j\) we obtain \[\begin{aligned} \mu^{*}(E)+\varepsilon &\geq\sum_{j=1}^{\infty}(\mu_0(B_j \cap C)+\mu_0(B_j \cap C^c))\\& \geq \sum_{j=1}^{\infty}(\mu^{*}(B_j \cap C)+\mu^{*}(B_j \cap C^c))\\&\geq (\mu^{*}(B \cap C)+\mu^{*}(B \cap C^c)) \\&\geq (\mu^{*}(E \cap C)+\mu^{*}(E \cap C^c)) \end{aligned}\] by subadditivity of \(\mu^*\). Letting \(\varepsilon \to 0\) we conclude \[\begin{aligned} \mu^{*}(E) \geq \mu^{*}(E \cap C)+\mu^{*}(E \cap C^c), \end{aligned}\] thus \(C\) is \(\mu^{*}\)-measurable and \(\mathcal A\subseteq \mathcal M(\mu^*)\) as desired.$$\tag*{$\blacksquare$}$$
If \(\nu\) is another measure on \(\mathcal M\) that extends \(\mu_0\), then \(\nu(E) \leq \mu(E)\) for all \(E \in \mathcal M\), with equality when \(\mu(E)<\infty\).
If \(\mu_0\) is \(\sigma\)-finite, then \(\mu\) is the unique extension of \(\mu_0\) to a measure on \(\mathcal M\).
Proof. For the first part of the theorem note that:
\((X, \mathcal M(\mu^*), \mu^*)\) is a complete measure space, which follows from the Carathéodory theorem.
\(\mathcal A\subseteq \mathcal M(\mu^*)\) by the previous proposition and consequently we obtain that \(\mathcal M=\sigma(\mathcal A)\subseteq \mathcal M(\mu^*)\).
Thus it suffices to define \(\mu\) to be the restriction of \(\mu^*\) to \(\mathcal M=\sigma(\mathcal A)\). Then we see that \(\mu(E)=\mu_0(E)\) for all \(E\in \mathcal A\).
Proof of (i). Assume that \(\nu\) is another measure on \(\mathcal M=\sigma(\mathcal A)\) such that \(\nu(E)=\mu_0(E)\) for all \(E \in \mathcal A\). We show that
\(\nu(E) \leq \mu(E)\) for all \(E \in \sigma(\mathcal A)\);
\(\nu(E)=\mu(E)\) if \(E \in \sigma(\mathcal A)\) and \(\mu(E)<\infty\).
For (a): If \(E \in \mathcal M\) and \(E \subseteq \bigcup_{j \in \mathbb{N}}A_j\), where \((A_j)_{j\in\mathbb N} \in \mathcal A\) then \[\nu(E) \leq \sum_{j=1}^{\infty}\nu(A_j)=\sum_{j=1}^{\infty}\mu_0(A_j)\] and we conclude that \({\color{red}\nu(E) \leq \mu(E)}\).
For (b): If we set \(A=\bigcup_{j=1}^{\infty}A_j\), then we have \[\color{blue} \nu(A)=\lim_{n \to \infty}\nu\big(\bigcup_{j=1}^{n}A_j\big) =\lim_{n \to \infty}\mu_0\big(\bigcup_{j=1}^{n}A_j\big)=\lim_{n \to \infty}\mu\big(\bigcup_{j=1}^{n}A_j\big)=\mu(A).\]
If \(\mu(E)<\infty\), let \(\varepsilon>0\) and choose \(A_j\)’s from \(\mathcal A\) such that \[\mu(A) \leq \mu(E)+\varepsilon,\] hence \(\mu(A \setminus E)<\varepsilon\) and \[\mu(E) \leq {\color{blue}\mu(A) =\nu(A)}=\nu(E)+{\color{red}\nu(A \setminus E)} \leq \nu(E)+{\color{red}\mu(A \setminus E)} \leq \nu(E)+\varepsilon.\]
Letting \(\varepsilon \to 0\) we see \(\mu(E) \leq \nu(E)\) if \(E \in \mathcal M=\sigma(\mathcal A)\) and \(\mu(E)<\infty\) and we conclude in this case that \[\nu(E)=\mu(E).\]Proof of (ii). Finally suppose that \(X=\bigcup_{j=1}^{\infty}A_j\), with \(\mu_0(A_j)<\infty\) and \(A_i \cap A_j=\varnothing\) if \(i \neq j\). Then for any \(E \in \mathcal M\) we have \[\mu(E)=\sum_{j=1}^{\infty}\mu(E \cap A_j)=\sum_{j=1}^{\infty}\nu(E \cap A_j)=\nu(E).\] This competes the proof of Carathéodory’s extension theorem. $$\tag*{$\blacksquare$}$$
Definition. Let \(X\neq\varnothing\) be a set and let \(\mathcal E\subseteq \mathcal P(X)\) be a collection containing \(\varnothing, X\). Let \(\rho:\mathcal E\to [0, \infty]\) be a set function such that \(\rho(\varnothing)=0\). For any \(A\subseteq X\)
\[\mu^{*}(A)=\inf\bigg\{\sum_{j=1}^{\infty}\rho(E_j):(E_j)_{j\in\mathbb N} \subseteq \mathcal{E} \text{ and } A \subseteq \bigcup_{j=1}^{\infty}E_j\bigg\}.\] defines an outer measure, which we call an outer measure induced by \(\rho\).
By Carathéodory’s theorem the collection of \(\mu^*\)-measurable sets \[\mathcal M(\mu^*) =\big\{A\subseteq X: \mu^{*}(E)=\mu^{*}(E \cap A)+\mu^{*}(E \cap A^c) \ \text{ for all } \ E \subseteq X\big\}.\] is a \(\sigma\)-algebra, which will be called the Carathéodory \(\sigma\)-algebra.
Moreover, \((X, \mathcal M(\mu^*), \mu^*)\) is a complete measure space, and the outer measure \(\mu^*\) restricted to the Carathéodory \(\sigma\)-algebra \(\mathcal M(\mu^*)\) will be called the Carathéodory measure.
\(\mathcal E_{\sigma}=\{\bigcup_{n\in\mathbb N}E_n: (E_n)_{n\in\mathbb N}\subseteq \mathcal E\}\) - the collection of countable unions of sets from \(\mathcal E\).
\(\mathcal E_{\delta}=\{\bigcap_{n\in\mathbb N}E_n: (E_n)_{n\in\mathbb N}\subseteq \mathcal E\}\) - the collection of countable intersections of sets from \(\mathcal E\).
\(\mathcal E_{\sigma\delta}=(\mathcal E_{\sigma})_\delta\) - the collection of countable intersections of sets from \(\mathcal E_\sigma\).
\(\mathcal E_{\delta\sigma}=(\mathcal E_{\delta})_\sigma\) - the collection of countable unions of sets from \(\mathcal E_\delta\).
\(\mathcal E^c=\{E^c: E\in \mathcal E\}\) - the collection of complements of sets from \(\mathcal E\).
Note that \((\mathcal E_{\sigma})^c=(\mathcal E^c)_{\delta}\) and \((\mathcal E_{\delta})^c=(\mathcal E^c)_{\sigma}\).
for any \(A \subseteq X\) and \(\varepsilon>0\) there exists a set \(U \supseteq A\) such that \[U\in \mathcal E_\sigma \quad\text{ and } \quad \mu^{*}(A) \leq \mu^{*}(U) \leq \mu^{*}(A)+\varepsilon.\]
for any \(A \subseteq X\) there exists a set \(B \supseteq A\) such that \[B\in \mathcal E_{\sigma\delta} \quad\text{ and } \quad \mu^{*}(A)=\mu^{*}(B).\]
Let \(A \subseteq X\), then by the definition of \(\mu^{*}\) for every \(\varepsilon>0\) there exists a sequence \((E_j)_{j=1}^{\infty}\subseteq \mathcal E\) such that \[A \subseteq \bigcup_{j \in \mathbb{N}}E_j=U\in \mathcal E_{\sigma} \qquad \text{ and } \qquad \sum_{j \in \mathbb{N}}\rho(E_j) \leq \mu^{*}(A)+\varepsilon.\] Since \(\mu^*=\rho\) on \(\mathcal E\) we obtain that \[\mu^{*}(A) \leq \mu^{*}(U)\le \sum_{j \in \mathbb{N}}\mu^*(E_j)= \sum_{j \in \mathbb{N}}\rho(E_j) \leq \mu^{*}(A)+\varepsilon.\]
Let \(A \subseteq X\), for every \(n \in \mathbb{N}\) there is a set \(U_n \in \mathcal E_{\sigma}\) such that \(A \subseteq U_n\) and \[\mu^{*}(A) \leq \mu^{*}(U_n) \leq \mu^{*}(A)+1/n, \qquad {\color{blue} \text{by (a) with }\ \varepsilon=1/n.}\] Define a set \(B=\bigcap_{n \in \mathbb{N}} U_{n}\in \mathcal E_{\sigma\delta}\) and note that \(A\subseteq B\) and \[\mu^{*}(A) \leq \mu^{*}(B) \leq \mu^{*}(U_n) \leq \mu^{*}(A)+1/{n},\] hence \(\mu^{*}(A)=\mu^{*}(B)\) as desired.$$\tag*{$\blacksquare$}$$
Then the outer measure is outer regular with respect to the family \(\mathcal E_{\sigma}\), which means that for every \(E\subseteq X\) one has \[\mu^*(E)=\inf\{\mu^*(U): U\supseteq E \text{ and } U\in\mathcal E_{\sigma}\}.\]
Let \[\lambda(E)=\inf\{\mu^*(U): U\supseteq E \text{ and } U\in\mathcal E_{\sigma}\}.\] we have to prove that \[\mu^*(E)=\lambda(E)\] for every \(E\in\mathcal P(X)\).
Fix \(E\in\mathcal P(X)\). Then for any \(U\in \mathcal E_{\sigma}\) such that \(E\subseteq U\) we have \(\mu^*(E)\le \mu^*(U)\), which yields that \(\mu^*(E)\le \lambda(E)\).
We now prove that \(\lambda(E)\le \mu^*(E)\). Let \(\varepsilon>0\), then by the previous proposition we can find \(U\in\mathcal E_{\sigma}\) such that \(U\supseteq E\) and \[\lambda(E)\le \mu^*(U)\le \mu^*(E)+\varepsilon.\] Since \(\varepsilon>0\) was arbitrary we obtain \[\lambda(E)\le \mu^*(E)\] as desired. $$\tag*{$\blacksquare$}$$
\(\mu^*=\rho\) on \(\mathcal E\);
\(\mathcal E\subseteq \mathcal M(\mu^*)\);
\(\mu^*\) is a \(\sigma\)-finite measure on \((X, \mathcal M(\mu^*))\).
\(E\in \mathcal M(\mu^*)\).
For every \(\varepsilon >0\) there exists a set \(U\supseteq E\) such that \[U\in \mathcal E_{\sigma} \quad\text{ and } \quad \mu^*(U\setminus E)\le \varepsilon.\]
There exists and a set \(G\supseteq E\) such that \[G\in\mathcal E_{\sigma\delta} \quad\text{ and } \quad \mu^*(G\setminus E)=0.\]
For every \(\varepsilon >0\) there exists and a set \(C\subseteq E\) such that \[C\in (\mathcal E^c)_{\delta} \quad\text{ and } \quad \mu^*(E\setminus C)\le \varepsilon.\]
There exists a set \(F\subseteq E\) such that \[F\in (\mathcal E^c)_{\delta\sigma} \quad\text{ and } \quad \mu^*(E\setminus F)=0.\]
Proof (i) \(\Longrightarrow\) (ii). Assume that \(E\in \mathcal M(\mu^*)\).
By the previous proposition for every \(\varepsilon>0\) there is a set \(U\supseteq E\) such that \(U\in \mathcal E_{\sigma}\) and \(\mu^{*}(E) \leq \mu^{*}(U) \leq \mu^{*}(E)+\varepsilon.\) If \(\mu^*(E)<\infty\) we obtain that \(\mu^{*}(U\setminus E)\le \varepsilon\), since by the Carathéodory condition \[\mu^{*}(E)+\mu^{*}(U\setminus E)=\mu^{*}(U\cap E)+\mu^{*}(U\cap E^c)=\mu^{*}(U) \leq \mu^{*}(E)+\varepsilon.\]
Assume now that \(\mu^*(E)=\infty\). Since \(\mu^*\) is \(\sigma\)-finite one can write \(X=\bigcup_{n\in \mathbb N}X_n\) for a sequence \((X_n)_{n\in \mathbb N}\subseteq \mathcal M(\mu^*)\) of disjoint sets such that \(\mu^*(X_n)<\infty\). Setting \(E_n=E\cap X_n\) for \(n\in\mathbb N\) one sees that \((E_n)_{n\in \mathbb N}\subseteq \mathcal M(\mu^*)\) and \(E=\bigcup_{n\in \mathbb N}E_n\) and \(\mu^*(E_n)<\infty\).
Applying the previous result to \(E_n\) for each \(n\in \mathbb N\) we obtain that for every \(\varepsilon>0\) there exists a set \(U_n\supseteq E_n\) such that \(U_n\in \mathcal E_{\sigma}\) and \[\mu^*(U_n\setminus E_n)\le \varepsilon2^{-n}.\]
We take \(U=\bigcup_{n\in \mathbb N}U_n\in \mathcal E_{\sigma}\) and note that \[\begin{aligned} \mu^*(U\setminus E)&=\mu^*\bigg(\bigcup_{n\in \mathbb N}U_n\cap \Big(\bigcup_{n\in \mathbb N}E_n\Big)^c\bigg)\\ &\le \sum_{n\in\mathbb N}\mu^*\bigg(U_n\cap \Big(\bigcup_{n\in \mathbb N}E_n\Big)^c\bigg)\\ &\le \sum_{n\in\mathbb N}\mu^*(U_n\setminus E_n)\\ &\le \sum_{n\in\mathbb N}\varepsilon2^{-n}\le \varepsilon, \end{aligned}\] since \(E_n\subseteq \bigcup_{n\in \mathbb N}E_n\) and \(\mu^*(U_n\setminus E_n)\le \varepsilon2^{-n}\). $$\tag*{$\blacksquare$}$$
Proof (ii) \(\Longrightarrow\) (iii). Assume that for every \(\varepsilon >0\) there exists a set \(U\supseteq E\) such that \(U\in \mathcal E_{\sigma}\) and \(\mu^*(U\setminus E)\le \varepsilon.\)
For each \(n\in \mathbb N\) with \(\varepsilon=1/n\) we find a set \(U_n\supseteq E\) such that \(U_n\in \mathcal E_{\sigma}\) and \(\mu^*(U_n\setminus E)\le 1/n\). Define a set \(G=\bigcap_{n\in \mathbb N}U_n\in \mathcal E_{\sigma\delta}\) then \(E\subseteq G\) and \(\mu^*(G\setminus E)\le \mu^*(U_n\setminus E)\le 1/n\). Thus \(\mu^*(G\setminus E)=0\). $$\tag*{$\blacksquare$}$$
Proof (iii) \(\Longrightarrow\) (i). Assume that there exists and a set \(G\supseteq E\) such that \(G\in \mathcal E_{\sigma\delta}\) and \(\mu^*(G\setminus E)=0\).
Since \(\mathcal E\subseteq \mathcal M(\mu^*)\) we have that \(\mathcal E_{\sigma\delta}\subseteq\sigma(\mathcal E)\subseteq \mathcal M(\mu^*)\) giving that \(G\in \mathcal M(\mu^*)\). Also \(G\setminus E\in \mathcal M(\mu^*)\) since \(\mu^*(G\setminus E)=0\) and \(\mu^*\) is complete on \((X, \mathcal M(\mu^*))\). Hence \(G\setminus(G\setminus E)\in \mathcal M(\mu^*)\), which implies \(E\in \mathcal M(\mu^*)\), since \(G\setminus(G\setminus E)=G\cap (G\cap E^c)^c=G\cap E=E\). $$\tag*{$\blacksquare$}$$
Proof (i) \(\Longrightarrow\) (iv). Assume that \(E\in \mathcal M(\mu^*)\).
Equivalently \(E^c\in \mathcal M(\mu^*)\), since (i) \(\Longleftrightarrow\) (ii) \(\Longleftrightarrow\) (iii), we may apply (ii) to \(E^c\). Then for every \(\varepsilon >0\) we find a set \(U\supseteq E^c\) such that \(U\in \mathcal E_{\sigma}\) and \(\mu^*(U\setminus E^c)\le \varepsilon\). But \(U\setminus E^c=E\setminus U^c\) so it suffices to take \(C=U^c\), then \(C\in (\mathcal E^c)_{\delta}\) and \(C\subseteq E\) and \(\mu^*(E\setminus C)\le \varepsilon\).$$\tag*{$\blacksquare$}$$
Proof (iv) \(\Longrightarrow\) (v). Assume that for every \(\varepsilon >0\) there exists a set \(C\subseteq E\) such that \(C\in (\mathcal E^c)_{\delta}\) and \(\mu^*(E\setminus C)\le \varepsilon.\)
For each \(n\in \mathbb N\) with \(\varepsilon=1/n\) we find a set \(C_n\subseteq E\) so that \(C_n\in (\mathcal E^c)_{\delta}\) and \(\mu^*(E\setminus C_n)\le 1/n\). Define a set \(F=\bigcup_{n\in \mathbb N}C_n\in (\mathcal E^c)_{\delta\sigma}\) then \(F\subseteq E\) and \(\mu^*(E\setminus F)\le \mu^*(E\setminus C_n)\le 1/n\). Thus \(\mu^*(E\setminus F)=0\). $$\tag*{$\blacksquare$}$$
Proof (v) \(\Longrightarrow\) (i). Assume that there exists and a set \(F\subseteq E\) such that \(F\in (\mathcal E^c)_{\delta\sigma}\) and \(\mu^*(E\setminus F)=0\).
Since \(\mathcal E\subseteq \mathcal M(\mu^*)\) we have that \((\mathcal E^c)_{\delta\sigma}\subseteq\sigma(\mathcal E)\subseteq \mathcal M(\mu^*)\) giving that \(F\in \mathcal M(\mu^*)\). Also \(E\setminus F\in \mathcal M(\mu^*)\) since \(\mu^*(E\setminus F)=0\) and \(\mu^*\) is complete on \((X, \mathcal M(\mu^*))\). Hence \(E=F\cup (E\setminus F)\in \mathcal M(\mu^*)\). $$\tag*{$\blacksquare$}$$
Remark
In fact we have proved that \[{\bf (i)\ \Longleftrightarrow \ (ii)\ \Longleftrightarrow \ (iii)} \qquad\text{ and } \qquad {\bf (i)\ \Longleftrightarrow \ (iv)\ \Longleftrightarrow \ (v).}\]
In practice, this theorem will be applied with various families \(\mathcal E\) of open sets, like for instance open intervals or open rectangles.
\(\mu^*\) is outer regular with respect to the family \(\mathcal E_{\sigma}\), i.e. \[\mu^*(E)=\inf\{\mu^*(U): U\supseteq E \text{ and } U\in\mathcal E_{\sigma}\} \ \text{ for any }\ E\in \mathcal M(\mu^*).\]
\(\mu^*\) is inner regular with respect to the family \((\mathcal E^c)_{\delta}\), i.e. \[\mu^*(E)=\sup\{\mu^*(C): C\subseteq E \text{ and } C\in(\mathcal E^c)_{\delta}\}\ \text{ for any }\ E\in \mathcal M(\mu^*).\ \]
To prove part (b) let \[\lambda(E)=\sup\{\mu^*(C): C\subseteq E \text{ and } C\in(\mathcal E^c)_{\delta}\}.\] we have to prove that \[\mu^*(E)=\lambda(E)\] for every \(E\in\mathcal M(\mu^*)\).
Fix \(E\in\mathcal M(\mu^*)\). Then for any \(C\in (\mathcal E^c)_{\delta}\) such that \(C\subseteq E\) we have \(\mu^*(C)\le \mu^*(E)\), which yields that \(\lambda(E)\le \mu^*(E)\).
We now prove that \(\mu^*(E)\le \lambda(E)\). Let \(\varepsilon>0\), then by the previous proposition we can find \(C\in(\mathcal E^c)_{\delta}\) such that \(C\subseteq E\) and \[\mu^*(E\setminus C)\le \varepsilon.\] This implies \(\mu^*(E)\le \lambda(E)\), since \(\varepsilon>0\) is arbitrary and \[\mu^*(E)=\mu^*(C)+ \mu^*(E\setminus C)\le \lambda(E)+ \varepsilon. \tag*{$\blacksquare$}\]
Fix \(E \in \mathcal M(\mu^*)\) and \(\varepsilon>0\). By the definition of \(\mu^{*}\) there is a sequence \((E_n)_{n \in \mathbb{N}}\subseteq\mathcal{E}\) so that \[E\subseteq \bigcup_{n \in \mathbb{N}}E_n \quad \text{ and } \quad \mu^{*}(E) \leq \sum_{n \in \mathbb{N}}\rho(E_n) \leq \mu^{*}(E)+\frac{\varepsilon}{2}.\]
Since \(\rho=\mu^*\) on \(\mathcal{E}\) thus \[\mu^{*}(E) \leq \sum_{n \in \mathbb{N}}\mu^*(E_n) \leq \mu^{*}(E)+\frac{\varepsilon}{2}.\]
For every \(n \in \mathbb{N}\) we have \(\bigcup_{k=1}^{n}E_k \in \mathcal{E}_{\sigma}\) and \(\mathcal{E}_{\sigma}\subseteq \sigma(\mathcal E)\subseteq \mathcal M(\mu^*)\), and consequently by the continuity of measure \[\lim_{n \to \infty}\mu^*\bigg(\bigcup_{k=1}^n E_k\bigg)=\mu^* \bigg(\bigcup_{n \in \mathbb{N}}E_n\bigg).\]
Since \(\mu^*\big(\bigcup_{n \in \mathbb{N}}E_n\big) \leq \mu^*(X)<\infty\) so there exist \(N \in \mathbb{N}\) so that \[\mu^*\bigg(\bigcup_{n \in \mathbb{N}}E_n\bigg) -\frac{\varepsilon}{2}<\mu^*\bigg(\bigcup_{n=1}^{N}E_n\bigg).\]
Then \[\mu^*\bigg(\bigcup_{n \in \mathbb{N}} E_n \setminus \bigcup_{k=1}^{N}E_k\bigg) =\mu^*\bigg(\bigcup_{n \in \mathbb{N}}E_n\bigg)-\mu^*\bigg(\bigcup_{k=1}^{N}E_k\bigg)<\frac{\varepsilon}{2}.\]
Let \(A = \bigcup_{n=1}^{N}E_n \in \mathcal{E}_{\sigma}\) and since \(E \subseteq \bigcup_{n \in \mathbb{N}}E_n\) we have \[\mu^*(E \setminus A) \leq \mu^* \bigg(\bigcup_{n \in \mathbb{N}}E_n \setminus A\bigg)<\frac{\varepsilon}{2}.\]
Similarly since \(A \subseteq \bigcup_{n \in \mathbb{N}}E_n\) and \(E \subseteq \bigcup_{n \in \mathbb{N}}E_n\) we have \[\begin{aligned} \mu^*(A \setminus E) &\leq \mu^* \left(\bigcup_{n \in \mathbb{N}}(E_n \setminus E)\right)\\ &=\mu^* \bigg(\bigcup_{n \in \mathbb{N}}E_n\bigg)-\mu^*(E) \\ &\leq \sum_{n \in \mathbb{N}}\mu^*(E_n)-\mu^*(E) \leq \frac{\varepsilon}{2}. \end{aligned}\]
Thus \[\mu^*(E \triangle A)=\mu^*(E \setminus A)+\mu^*(A \setminus E)<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon.\] as desired. $$\tag*{$\blacksquare$}$$