A set $S\subseteq X$ is $\mu$ -measurable if and only if \begin{equation}\label{eq:mu-measurable inequal} \mu(E)\ge\mu(E\cap S)+\mu(E\cap S^c) \end{equation}for every $E\subseteq X$ . As this inequality is clearly satisfied if $S=\emptyset$ and is unchanged when $S$ is replaced by $S^c$ , then $\mathcal{A}$ contains the empty set and is closed under taking complements of sets. To show that $\mathcal{A}$ is a $\sigma$ -algebra, it only remains to show that it is closed under taking countable unions of sets. Choose any sets $A,B\in\mathcal{A}$ and $E\subseteq X$ . Then, \begin{equation*}\begin{split} \mu(E)&\ge\mu(E\cap A) + \mu(E\cap A^c)\\ &\ge\mu(E\cap A) + \mu(E\cap A^c\cap B) + \mu(E\cap A^c\cap B^c)\\ &\ge\mu(E\cap(A\cup B))+\mu(E\cap A^c\cap B^c) \end{split}\end{equation*}The first two inequalities here follow from applying () with $A$ and then $B$ in place of $S$ , and the third uses the subadditivity of $\mu$ together with $A\cup (A^c\cap B) = A\cup B$ . So () is satisfied with $A\cup B$ in place of $S$ , showing that $\mathcal{A}$ is closed under taking pairwise unions and is therefore an algebra of sets on $X$ . If $A,B$ are disjoint sets in $\mathcal{A}$ then replaceing $E$ by $A\cap B$ and $S$ by $A$ in () gives $\mu(A\cap B)\ge\mu(A)+\mu(B)$ . As the reverse inequality follows from subadditivity of $\mu$ , this implies that $\mu$ is an additive set function on $\mathcal{A}$ .

Now choose a sequence $A_i\in\mathcal{A}$ , and set $B_i\equiv\bigcup_{j=1}^{i}A_j$ which are in the algebra $\mathcal{A}$ . To prove that $\mathcal{A}$ is a $\sigma$ -algebra it needs to be shown that $A\equiv\bigcup_iA_i=\bigcup_iB_i$ is itself in $\mathcal{A}$ . First, as $B_i\in\mathcal{A}$ and $A^c\subseteq B_i^c$ , \begin{equation*} \mu(E)\ge\mu(E\cap B_i)+\mu(E\cap B_i^c)\ge\mu(E\cap B_i)+\mu(E\cap A^c). \end{equation*}As $C_i\equiv B_i\setminus B_{i-1}$ are pairwise disjoint sets in $\mathcal{A}$ satisfying $\bigcup_{j=1}^i C_j=B_i$ the additivity of $\mu$ gives \begin{equation*} \mu(E)\ge\sum_{j=1}^i\mu(E\cap C_j)+\mu(E\cap A^c). \end{equation*}So, letting $i$ increase to infinity, the subadditivity of $\mu$ applied to $\bigcup_j(E\cap C_j)=E\cap A$ gives \begin{equation*} \mu(E)\ge\sum_j\mu(E\cap C_j)+\mu(E\cap A^c)\ge\mu(E\cap A)+\mu(E\cap A^c). \end{equation*}This shows that $A$ is $\mu$ -measurable and so $\mathcal{A}$ is a $\sigma$ -algebra.

It only remains to show that the restriction of $\mu$ to $\mathcal{A}$ is a measure, for which it needs to be shown that $\mu$ is countably additive on $\mathcal{A}$ . So, choose any pairwise disjoint sequence $A_i\in\mathcal{A}$ and set $A=\bigcup_iA_i$ . The following inequality \begin{equation*} \sum_{j=1}^i\mu(A_j)=\mu\left(\bigcup_{j=1}^iA_j\right)\le\mu(A)\le\sum_j\mu(A_j) \end{equation*}follows from the additivity of $\mu$ on $\mathcal{A}$ , the requirement that $\mu$ is increasing and from the countable subadditivity of $\mu$ . Letting $i$ increase to infinity gives $\mu(A)=\sum_j\mu(A_j)$ and $\mu$ is indeed countably additive on $\mathcal{A}$ .