> [!definition] > > Let $(X, \cm)$ be a [[Measure Space|measurable space]] and $\mu, \nu$ be [[Signed Measure|signed]] or positive measures. $\mu$ and $\nu$ are **mutually singular** ($\mu \perp \nu$) if there exists a partition $U, V$ of $X$ such that $U$ is $\nu$-null and $V$ is $\mu$-null. > [!theorem] > > Let $(X, \cm)$ be a measurable space and $\mu, \nu$ be finite measures. Then they are either mutually singular or > $ > \exists \varepsilon > 0, E \in \cm: \mu(E) > 0, \nu \ge \varepsilon\mu \text{ on }E > $ > There exists $\varepsilon > 0$, $E \in \cm$ such that $\mu(E) > 0$ and $E$ is a positive set for $\nu - \varepsilon \mu$. > > *Proof*. Let $P_n \cup N_n$ be a [[Hahn Decomposition Theorem|Hahn decomposition]] for $\nu - \mu/n$, and take $P = \bigcup_{n \in \nat}P_n$ and $N = \bigcap_{n \in \nat}N_n = P^c$. Then $N$ is a negative for $v - \mu/n$, for all $n \in \nat$, meaning that $0 \le \nu(N) \le \mu(N)/n$ for all $n \in \nat$ and $\nu(N) = 0$. If $\mu(P) = 0$, then $\nu \perp \mu$. > > If $\mu(P) > 0$, then $\mu(P) = \limv{n}\mu(P_n)$ by continuity from below and there exists $n \in \nat$ such that $\mu(P_n) > 0$, and $P_n$ is a positive set for $\nu - \mu/n$. > [!theorem] > > Let $(X, \cm, \mu)$ be a measure space and $\seq{\nu_n}$ be a sequence of measures on $(X, \cm)$. If $v_n \perp \mu$ for all $n \in \nat$, then $\sum_{n \in \nat}v_n \perp \mu$. > > *Proof*. Let $X = U_n \cup V_n$ such that $V_n$ is $\mu$-null and $U_n$ is $\nu_n$-null. Let $U = \bigcap_{n \in \nat}U_n$ and $V = \bigcup_{n \in \nat}V_n = U^c$. Then $V$ is still $\mu$-null and $U$ is $v_n$-null for all $n \in \nat$. Therefore $U$ is null for $\sum_{n \in \nat}\nu_n$, and $\sum_{n \in \nat}v_n \perp \mu$.