> [!definition] > > Let $(X, \cm, \mu)$ be a [[Measure Space|measure space]]. $\mu$ is **semifinite** if > $ > \forall E \in \cm: \mu(E) = \infty, \exists F \subset E: 0 < \mu(F) < \infty > $ > [!theorem] > > Let $\mu$ be a semifinite measure and $E \in \cm$, define > $ > \cf_E = \bracs{B \subset E: \mu(B) < \infty} > $ > then $\sup_{F \in \cf(E)}\mu(F) = \mu(E)$. > > *Proof*. If $\mu(E) < \infty$, then $E$ itself satisfies the criterion. > > Now suppose that $\mu(E) = \infty$. Suppose for the sake of contradiction that $M = \sup_{F \in \cf(E)}\mu(F) < \infty$. Let $\seq{F_n}$ such that $\mu(F_n) \upto M$, then > $ > \mu\paren{\bigcup_{n \in \nat}F_n} = \limv{n}\mu\paren{\bigcup_{k = 1}^{n}F_n} = M > $ > Let $F = \bigcup_{n \in \nat}F_n$, then since $\mu(E) = \infty$, $\mu(E \setminus F) = \infty$, and there exists $G \subset E \setminus F: 0 < \mu(G) < \infty$. This gives $F \cup G \subset E$ and $M < \mu(F \cup G) < \infty$, which contradicts the fact that $M$ is the supremum. > [!theorem] > > Let $(X, \cm, \mu)$ be a measure space and let > $ > \mu_0(E) = \sup\bracs{\mu(F): F \subset E, \mu(F) < \infty} > $ > then $\mu_0$ is a semifinite measure, known as the **semifinite part** of $\mu$. > > *Proof*. > > ### Basic Properties > > Let $E \in \cm$ such that $\mu(E) < \infty$, then $\mu_0(E) = \mu(E)$. > > *Proof*. By monotonicity, $\mu(F) \le \mu(E)$ for all $F \subset E$. This means that $\mu_0(E) \le \mu(E)$. Since $E$ itself is finite, $\mu_0(E) \ge \mu(E)$. > > > ### Measure > > $\mu_0$ is a measure. > > *Proof*. Firstly, $\mu_0(\emptyset) = \mu(\emptyset) = 0$. > > Now, let $\seq{E_n}$ be a [[Sequence|sequence]] of disjoint sets. If there exists $n \in \nat$ such that $\mu_0(E_n) = \infty$, then > $ > \bracs{F \subset E_n: \mu(F) < \infty} \subset \bracs{F \subset \bigcup_{n \in \nat}E_n: \mu(F) < \infty} > $ > Therefore $\mu_0(E) \ge \mu_0(E_n) = \infty$, and $\mu_0(E) = \infty$. > > Now suppose that $\mu_0(E_n) < \infty$ for all $n \in \nat$ and $\mu_0\paren{\bigcup_{n \in \nat}E_n} < \infty$. Let $\varepsilon > 0$ and $F_n \subset E_n$ such that $\mu(F_n) > \mu_0(E_n) - \varepsilon/2^n$. This means that > $ > \begin{align*} > \mu_0\paren{\bigcup_{n \in \nat}E_n} &\ge \mu\paren{\bigcup_{k \le n}F_k} = \sum_{k \le n}\mu(F_k) \quad \forall n \in \nat \\ > \mu_0\paren{\bigcup_{n \in \nat}E_n} &\ge \sum_{n \in \nat}\mu(F_n) > \sum_{n \in \nat}\braks{\mu_0(E_n) - \varepsilon/2^n} \\ > \mu_0\paren{\bigcup_{n \in \nat}E_n} &\ge \sum_{n \in \nat}\mu_0(E_n) - \varepsilon > \end{align*} > $ > Since the above applies to all $\varepsilon > 0$, $\mu_0\paren{\bigcup_{n \in \nat}E_n} \ge \sum_{n \in \nat}\mu_0(E_n)$. > > Let $F \subset \bigcup_{n \in \nat}E_n$ such that $\mu(F) < \infty$, then > $ > \mu(F) \le \sum_{n \in \nat}\mu(F \cap E_n) \le \sum_{n \in \nat}\mu_0(E_n) > $ > Since this applies to all $F \subset \bigcup_{n \in \nat}E_n$, $\mu_0\paren{\bigcup_{n \in \nat}E_n} \le \sum_{n \in \nat}\mu_0(E_n)$. > > Finally, suppose that $\mu_0(E_n) < \infty$ for all $n \in \nat$, and $\mu_0\paren{\bigcup_{n \in \nat}E_n} = \infty$. Let $\alpha \in \real$ and $F \subset \bigcup_{n \in \nat}E_n$ such that $\alpha < \mu(F) < \infty$, then > $ > \sum_{n \in \nat}\mu_0(E_n \cap F) \ge \sum_{n \in \nat}\mu(E_n \cap F) = \mu(F) > \alpha > $ > Therefore $\sum_{n \in \nat}\mu_0(E_n \cap F) = \mu_0\paren{\bigcup_{n \in \nat}E_n}$. > > ### Semifinite > > $\mu_0$ is semifinite. > > *Proof*. Let $E \in \cm: \mu_0(E) \ne 0$, then > $ > \mu_0(E) = \sup\bracs{\mu(F): F \subset E, \mu(F) < \infty} > $ > There exists $F \subset E$ such that $0 < \mu(F) < \infty$.