> [!definition] > > Let $(X, \cm, \mu)$ be a finite [[Measure Space|measure space]]. For each $E, F \in \cm$, define > $ > d(E, F) = \mu(E \Delta F) > $ > then $d$ is a pseudo-[[Metric Space|metric]] on $\cm$, known as the **Fréchet-Nikodym metric**. Define $E \sim F$ if $d(E, F) = 0$, then $\cm/\sim$ is [[Complete Metric Space|complete]] with respect to $d$, denoted as $\cm_0$. > > *Proof*. Let $\seq{E_n} \subset \cm$ be a [[Cauchy Sequence|Cauchy sequence]] in $\cm$. By completeness of $L^1(\mu)$, there exists $f \in L^1(\mu)$ such that $\one_{E_{n_k}} \to f$ in $L^1$. Assume without loss of generality that $\one_{E_{n_k}} \to f$ [[Almost Everywhere|a.e.]] as well, then $f$ there must exist $E \in \cm$ such that $f = \one_E$. Therefore $E_n \to E$ with respect to $d$. > [!theorem] > > Let $(X, \cm, \mu)$ be a finite [[Measure Space|measure space]]. Then > $ > \mu(E \Delta F) = 0 \Rightarrow \mu(E) = \mu(F) > $ > *Proof*. > $ > \mu(E \Delta F) = 0 \Rightarrow \mu(E \setminus F) = \mu(F \setminus E) = 0 > $ > Now, > $ > \mu(E \cup F) = \mu(E) + \mu(F \setminus E) = \mu(E) = \mu(F) > $ > [!theorem] > > Let $(X, \cm, \mu)$ be a finite measure space. Define a [[Relation|relation]] $\sim$ on $\cm$ where $E \sim F \Leftrightarrow \mu(E) = \mu(F) \Leftrightarrow \mu(E \Delta F) = 0$. Then $\sim$ is an [[Equivalence Relation|equivalence relation]]. > > *Proof*. Firstly, $\mu(E \Delta E) = \mu(\emptyset) = 0$ and $\sim$ is reflexive. Then, $\mu(E \Delta F) = \mu(F \Delta E)$, so $E \sim F \Leftrightarrow F \sim E$ and $\sim$ is symmetric. > > Finally, $E \sim F$, $F \sim G$ implies that $\mu(E) = \mu(F) = \mu(G)$ and $E \sim G$, $\sim$ is transitive. > [!definition] > > Let $(X, \cm, \mu)$ be a measure space and $\rho(E, F) = \mu(E \Delta F)$, then $\rho$ is a [[Metric Space|metric]] on the space $\cm/\sim$ of [[Equivalence Class|equivalence classes]]. > > *Proof*. Let $E, F, G \in \cm$, then > $ > (E \Delta F) \cup (F \Delta G) \supseteq E \Delta G > $ > Because $G \setminus E \subseteq (F \setminus E) \cup (G \setminus F)$ ($F \setminus E$ fills in places removed by $F$) and the other way around. Therefore $\rho$ satisfies the triangle inequality. > > $\rho$ is definitely non-negative, and since we defined an equivalence relation, any two members of $\cm$ having a distance of $0$ must be in the same equivalence class.