> [!definition] > > Let $(X, \cm, \mu)$ be a [[Measure Space|measure space]] and $\nu$ be a [[Signed Measure|signed measure]] on $(X, \cm)$. The signed measure $\nu$ is **absolutely continuous** with respect to $\mu$ (denoted as $\nu \ll \mu$) if $\nu(E) = 0$ whenever $\mu(E) = 0$. > [!theorem] > > Let $(X, \cm, \mu)$ be a measure space, $\nu$ be a signed measure on $(X, \cm)$, and $\nu = \nu^+ - \nu^-$ be its [[Jordan Decomposition Theorem|Jordan decomposition]]. Then $\nu \ll \mu$ if and only if $\nu^+ \ll \mu$ and $\nu^- \ll \mu$. > > *Proof*. Suppose that $\nu \ll \mu$ and $\mu(E) = 0$. Let $X = P \cup N$ be a [[Hahn Decomposition Theorem|Hahn decomposition]] for $\nu$, then $\mu(E \cap P) = \mu(E \cap N) = 0$ as well. Therefore $\nu(E \cap P) = \nu^+(E) = 0$ and $\nu(E \cap N) = -\nu^-(E) = 0$ and $\nu^+ \ll \mu$, $\nu^- \ll \mu$. > [!theorem] > > Let $(X, \cm, \mu)$ be a measure space and $\nu$ a finite signed measure on $(X, \cm)$. Then $\nu \ll \mu$ if and only if > $ > \forall \varepsilon > 0, \exists \delta > 0: \mu(E) < \delta \Rightarrow |\nu(E)| < \varepsilon > $ > *Proof.* Suppose that > $ > \forall \varepsilon > 0, \exists \delta > 0: \mu(E) < \delta \Rightarrow \nu(E) < \varepsilon > $ > Let $E \in \cm: \mu(E) = 0$ be a $\mu$-[[Null Set|null set]], then $\mu(E) < \delta$ for all $\delta > 0$. Then for all $\varepsilon > 0$, $|\nu(E)| < \varepsilon$ and $|\nu(E)| = 0$. Therefore $\nu \ll \mu$. > > Since $\nu \ll \mu$ if and only if $|\nu| \ll \mu$, and that $|\nu(E)| \le |\nu|(E)$, we can assume that $\nu$ is also positive. Now suppose that the $\varepsilon$-$\delta$ condition does not hold, > $ > \exists \varepsilon > 0: \forall \delta > 0, \exists E \in \cm: \mu(E) < \delta, |\nu(E)| \ge \varepsilon > $ > Take $E_n \in \cm$ such that $\mu(E_n) < 2^{-n}$ but $\nu(E) \ge \varepsilon$. Let $F_n = \bigcup_{k = n}^{\infty}E_k$, then $\mu(F_n) < \sum_{k = n}^{\infty}2^{-k} = 2^{-(k-1)}$ . Let $F = \bigcap_{n \in \nat}F_k$, then by continuity from above, as $\nu$ is finite, > $ > \mu(F) = \limv{n}\mu(F_n) = 0 \quad \nu(F) = \limv{n}\nu(F_n) \ge \varepsilon > $ > and $\nu \not\ll \mu$. > [!theorem] > > Let $(X, \cm, \mu)$ be a measure space and $f: X \to \real$ be an extended $\mu$-[[Integrable Function|integrable function]]. Then the signed measure $\nu(E) = \int_E fd\mu$ is is absolutely continuous with respect to $\mu$. $\nu$ is finite if and only if $f \in L^1(\mu)$. > > Moreover, if $f \in L^1(\mu)$, then > $ > \forall \varepsilon > 0, \exists \delta > 0: \mu(E) < \delta \Rightarrow \abs{\int_E f d\mu} < \varepsilon > $ > and $\nu(E) = \int_E f d\mu$ can be denoted as $d\nu = f d\mu$. > [!theorem] > > Let $(X, \cm, \mu)$ be a measure space and $\seq{\nu_n}$ be a [[Sequence|sequence]] of measures on $(X, \cm)$. If $v_n \ll \mu$ for all $n \in \nat$, then $\sum_{n \in \nat}v_n \ll \mu$. > > *Proof*. Let $E \in \cm: \mu(E) = 0$. Then $\nu_n(E) = 0 \forall n \in \nat$, and $\sum_{n \in \nat}\nu_n(E) = 0$. Therefore $\sum_{n \in \nat}\nu_n \ll \mu$.