> [!theoremb] Theorem > > Let $X$ be a [[Set|set]], $\alg \subseteq \pow{X}$ be an [[Algebra|algebra]], $\mu_0$ be a [[Premeasure|premeasure]] on $\alg$, and $\cm = \cm(\alg)$ be the [[Sigma Algebra|sigma algebra]] generated by $\alg$. > > Then there exists a [[Measure Space|measure]] on $\cm$ whose restriction to $\alg$ is $\mu_0$, being the induced [[Outer Measure|outer measure]] $\mu = \mu^*$. > > If $v$ is another measure on $\cm$ that extends $\mu_0$, then $v(E) \le \mu(E)$ for any $E \in \cm$, 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 $\cm$. > > *Proof*. By [[Carathéodory's Theorem]], the induced outer measure $\mu^*$ has a collection of measurable sets $M$ being a sigma algebra. Since $\alg \subseteq M$ are all outer measurable, $M \supseteq \cm(\alg)$ contains a "spanning set" of $\cm(\alg)$, and therefore contains $\cm = \cm{(\alg)}$. As shown earlier, the restriction of $\mu^*$ to $\alg$ is equal to $\mu_0$. > > Let $v$ be another measure on $\cm$ that extends $\mu_0$, $E \in \cm$, and $\bracs{E_i}_1^{\infty} \supseteq E$ be an elementary cover of $E$. Then > $ > v(E) = v\paren{\bigcup_{i \in \nat}E_i} \le \sum_{i \in \nat}v(E_i) = \sum_{i \in \nat}\mu_0(E_i) > $ > and $v(E)$ is a lower bound for > $ > \bracs{ > \sum_{i \in I}\mu_0(E_i): |I| = |\nat|, E_i \in \alg, \bigcup_{i \in \nat}E_i \supseteq E > } > $ > and we have $v(E) \le \mu^*(E)$ since $\mu^*(E)$ is the greatest lower bound. > > We take advantage of the fact that $v$ is also an extension of $\mu_0$ by taking limits on sets that they agree on. Let $\bracs{E_i}_1^{\infty}$ be an elementary cover of $E$, and take > $ > A_j = \bigcup_{i = 1}^{j}E_i \quad A_j \in \alg > $ > then > $ > v(A_j) = \mu^*(A_j) = \mu_0(A_j) \quad \forall j \in \nat > $ > and using continuity from below, > $ > v\paren{\bigcup_{i \in \nat}E_i} = \limv{i}v(A_i) = \limv{i}\mu^*(A_i) = \mu^*\paren{\bigcup_{i \in \nat}E_i} > $ > Suppose that $\mu^*(E) < \infty$, then since $\mu^*$ is the infimum, > $ > \forall \varepsilon > 0, \exists I: |I| = |\nat|, E_i \in \alg, \bigcup_{i \in I}E_i \supseteq E, \sum_{i \in I}\mu_0(E_i) < \mu^*(E) + \varepsilon > $ > Take $\bracs{E_i}_1^{\infty}$ such that > $ > \mu^*(E) + \varepsilon > \sum_{i \in \nat}\mu_0(E_i) \ge \mu^*\paren{\bigcup_{i \in \nat}E_i} = v\paren{\bigcup_{i \in \nat}E_i} \ge \mu^*(E) > $ > with the inequality coming from subadditivity and monotonicity. Decompose > $ > \begin{align*} > \mu^*\paren{\bigcup_{i \in \nat}E_i} &= \mu^*(E) + \mu^*\paren{\bigcup_{i \in \nat}E_i \setminus E} < \mu^*(E) + \varepsilon\\ > \mu^*\paren{\bigcup_{i \in \nat}E_i \setminus E} &< \varepsilon\\ > v\paren{\bigcup_{i \in \nat}E_i} &= v(E) + v\paren{\bigcup_{i \in \nat}E_i \setminus E} \\ > v\paren{\bigcup_{i \in \nat}E_i \setminus E} &\le > \mu^*\paren{\bigcup_{i \in \nat}E_i \setminus E} < \varepsilon > \end{align*} > $ > and we obtain > $ > \mu^*(E) \le v\paren{\bigcup_{i \in \nat}E_i} = v(E) + v\paren{\bigcup_{i \in \nat}E_i \setminus E} < v(E) + \varepsilon > $ > Rearranging the equation, and since we can find an elementary cover for any $\varepsilon > 0$, > $ > \mu^*(E) - \varepsilon < \mu^*(E) - v\paren{\bigcup_{i \in \nat}E_i \setminus E} < v(E) \le \mu^*(E) > $ > we can approach $v(E)$ from below by arbitrarily squishing the overcover. This is possible because $\cm$ is a sigma algebra generated by $\alg$, and allows us to approach $E$ and its measure using just the elementary sets. > > Let $\varepsilon_n = \frac{1}{n}$ and $I_n$ be such that $\bigcup_{i \in I_n}E_i \supseteq E, E_i \in \alg$ and $\sum_{i \in I_n}\mu_0(E_i) < \mu^*(E) + \varepsilon_n$. This gives > $ > x_n = \mu^*(E) - v\paren{\bigcup_{i \in I_n}E_i \setminus E} > \mu^*(E) - \frac{1}{n} > $ > then > $ > x_n < v(E) \le \mu^*(E) \forall n \in \nat \quad \limv{n}x_n = \mu^*(E) > $ > which gives > $ > v(E) = \mu^*(E) > $ > > Suppose that $\mu^*$ is $\sigma$-finite, then we can decompose > $ > X = \bigcup_{j \in \nat}E_j \quad \mu^*(E_j) < \infty > $ > further into > $ > F_j = E_j \setminus \bigcup_{i = 1}^{j - 1}E_i \quad \bigcup_{i \in \nat}F_j = X \quad v(E_j) \le \mu^*(E_j) < \infty > $ > and the $F_j$s form a disjoint union. > > Then taking any $E \in \cm$, > $ > \begin{align*} > \mu^*(E) &= \mu^*\paren{\bigcup_{i \in \nat}E \cap F_i} \\ > &= \sum_{i \in \nat}\mu^*(E \cap F_i) \\ > &= \sum_{i \in \nat}v(E \cap F_i) \\ > &= v\paren{\bigcup_{i \in \nat}E \cap F_i} \\ > &= v(E) > \end{align*} > $ > and $\mu^* = v$ is unique.