Proof. Let

then $\mu^{*}$ is an outer measure by Proposition 14.7.2. By Carathéodory’s theorem, there exists a measure space $(X, \cm, \mu)$ such that $\mu = \mu^{*}|_{\cm}$.

(1): Let $E \in \alg$ and $F \subset X$. If there exists $\seq{F_n}\subset \alg$ with $\bigcup_{n \in \natp}F_{n} \supset F$, then

for all such $\seq{F_n}$, so $\mu^{*}(F) = \mu^{*}(F \cap E) + \mu^{*}(F \setminus E)$. Otherwise, $\mu^{*}(F) = \infty$ and the result holds directly.

(2): Let $E \in \alg$ and $\seq{E_n}\subset \alg$ such that $\bigcup_{n \in \natp}E_{n} \supset E$, then

As this holds for all such $\seq{E_n}$, $\mu^{*}(E) = \mu_{0}(E)$.

(U): Let $\seq{E_n}\subset \alg$, then by continuity from below (Proposition 14.1.5),

Thus if $E \in \cm \cap \cn$ with $\bigcup_{n \in \natp}E_{n} \supset E$, then

As this holds for all such $\seq{E_n}$, $\nu(E) \le \mu^{*}(E) = \mu(E)$.

Suppose that $\mu(E) < \infty$, then there exists $\seq{E_n}\subset \alg$ such that $\bigcup_{n \in \natp}E_{n} \supset E$ and $\sum_{n \in \natp}\mu_{0}(E_{n}) < \infty$. In which case,

so $\nu(E) = \mu(E)$ whenever $\mu(E) < \infty$.

Finally, if $\mu$ is $\sigma$-finite, then there exists $\seq{E_n}\subset \cm$ such that $E_{n} \upto X$ and $\mu(E_{n}) < \infty$ for all $n \in \natp$. For each $n \in \natp$, there exists $\bracsn{E_{n, k}}_{k = 1}^{\infty} \subset \alg$ such that $E_{n} \subset \bigcup_{k \in \natp}E_{n, k}$ and $\mu\paren{\bigcup_{k \in \natp}E_{n, k}}< \infty$. Let $F_{n} = \bigcup_{j = 1}^{n} \bigcup_{k \in \natp}E_{j, k}$, then $\seq{F_n}\subset \sigma(\alg) \subset \cm \cap \cn$ with $F_{n} \upto X$ and $\mu(F_{n}) < \infty$ for all $n \in \natp$. For any $E \in \cm \cap \cn$, by continuity from below (Proposition 14.1.5),

so $\nu|_{\cm \cap \cn}= \mu|_{\cm \cap \cn}$.$\square$