> [!definition] > > Let $(X, \cm)$ be a [[Measure Space|measurable space]]. A **signed measure** on $(X, \cm)$ is a [[Function|function]] $\nu: \cm \to [-\infty, \infty]$ such that > - $\nu(\emptyset) = 0$ > - $\nu$ assumes at most one of the values $\pm \infty$. > - For any [[Sequence|sequence]] $\seq{E_j} \subset \cm$ of disjoint sets, $\nu\paren{\bigcup_{i = 1}^{\infty}E_j} = \sum_{i = 1}^{\infty}\nu(E_j)$. Where the latter sum [[Limit|converges]] absolutely if $\nu\paren{\bigcup_{i = 1}^{\infty}E_j}$ is finite. > > *Proof*. The equality in the third condition is delicate, as while the union is always the same regardless of the order, the sum may be different. Essentially, this requires the sum to have a consistent value regardless of the arrangement of its terms. > > This problem boils down to a few cases, related to [[Riemann's Rearrangement Theorem]]. Let $\seq{\nu(E_j)}$ be a sequence and let $\bracs{\nu(P_{j})}$ be the non-negative terms of the sum, and $\bracs{\nu(Q_j)}$ be the negative terms of the sum. For the sum to be consistent regardless of the order, one of $\sum_{j \in \nat}\nu(P_j)$ and $\sum_{j \in \nat}\nu(Q_j)$ has to be finite. If the other one diverges to $\pm \infty$, then the sum is also $\pm \infty$, regardless of order. If both series converge, then the sum converges absolutely, and to one value regardless of order. > [!theorem] > > Let $\mu_1, \mu_2$ be measures on $(X, \cm)$ where one of them is finite, then $\nu = \mu_1 - \mu_2$ is a signed measure. > > Let $(X, \cm, \mu)$ be a measure space and $f: X \to \real$ be a $\cm$-[[Measurable Function|measurable function]] where $\int f^+$ is finite or $\int f^-$ is finite, then $\nu(E) = \int_{E}fd\mu$ is a measure. > > *Proof*. Firstly, $\nu(\emptyset) = \mu_1(\emptyset) - \mu_2(\emptyset) = 0$. Since one of them is finite, $\nu$ can only assume one of $\pm\infty$. > > Now, let $\seq{E_n} \subset \cm$ be a sequence of disjoint sets, then since one of $\mu_1$ and $\mu_2$ is finite, $\sum_{n \in \nat: \nu(E_n) \ge 0}\nu(E_n)$ or $\sum_{n \in \nat: \nu(E_n) < 0}\nu(E_n)$ (or both) is finite. Meaning that the sum $\sum_{n = 1}^{\infty}\nu(E_n)$ converges to the same value regardless of rearrangements. This way, > $ > \begin{align*} > \nu\paren{\bigcup_{n \in \nat}E_n} &= \mu_1\paren{\bigcup_{n \in \nat}E_n} - \mu_2\paren{\bigcup_{n \in \nat}E_n} \\ > &= \sum_{n \in \nat}\mu_1(E_n) - \sum_{n \in \nat}\mu_2(E_n) \\ > &= \sum_{n = 1}^{\infty}\mu_1(E_n) - \mu_2(E_n) \\ > &= \sum_{n = 1}^{\infty}\nu(E_n) > \end{align*} > $ > If $\nu\paren{\bigcup_{n \in \nat}E_n} < \infty$, then $\sum_{n \in \nat}\mu_1(E_n)$ is finite and the series must converge absolutely. Therefore $\nu$ is a signed measure. > > Let $(X, \cm, \nu)$ be a measure space and $f: X \to \real$ be a $\cm$-measurable function where $\int f^+$ or $\int f^-$ is finite. Let $\mu_1(E) = \int_E f^+$, $\mu_2(E) = \int_E f^-$, then one of $\mu_1$ and $\mu_2$ is finite, meaning that > $ > \nu(E) = \int_E f = \int_E f^+ - \int_E f^- > $ > is a signed measure. > [!theorem] > > Let $(X, \cm)$ be a measurable space and $\nu$ be a signed measure on $\cm$. Let $\seq{E_n} \subset \cm$ be an ascending sequence of sets, then > $ > \nu\paren{\bigcup_{n \in \nat}E_n} = \limv{n}\nu(E_n) > $ > Let $\seq{F_n} \subset \cm$ be a descending sequence of sets and $\nu(F_1) \in \real$, then > $ > \paren{\bigcap_{n \in \nat}F_n} = \limv{n}\nu(F_n) > $ > [!definition] > > Let $(X, \cm)$ be a measurable space and $\nu$ be a signed measure. A set $E \in \cm$ is **positive/negative/[[Null Set|null]]** if $\nu(F)$ is non-negative/non-positive/zero for any $F \in \cm, F \subseteq E$. > [!theorem] > > Let $(X, \cm)$ be a measurable space and $\nu$ be a signed measure. > > If $E \in \cm$ is positive/negative/null, then any measurable subset of $E$ is also positive/negative/null. > > If $\seq{E_n} \subset \cm$ is a sequence of positive/negative/null sets, then their union is also positive/negative/null.