> [!definition] > > Let $(X, \cm)$ be a [[Measure Space|measurable space]], a [[Function|function]] $\nu: \cm \to \complex$ is a **complex measure** if > - $\nu(\emptyset) = 0$ > - If $\seq{E_n} \subset \cm$ is a [[Sequence|sequence]] of disjoint sets, then $\nu\paren{\bigcup_{n \in \nat}E_n} = \sum_{n \in \nat}\nu(E_n)$, where the latter sum converges absolutely (see [[Riemann's Rearrangement Theorem]]). > [!definition] > > Let $\nu$ be a complex measure, then $\nu_r$ and $\nu_i$ be its real and imaginary parts. > > If $\mu$ is a positive measure, then $\nu$ is [[Mutual Singularity|mutually singular]] with respect to $\mu$ if and only if $\nu_r \perp \mu$ and $\nu_i \perp \mu$, and $\nu$ is [[Absolute Continuity|absolutely continuous]] with respect to $\mu$ if and only if $\nu_r, \nu_i \ll \mu$. > [!theorem] > > Let $\nu$ be a complex measure, then the [[Integrable Function|space of integrable functions]] is defined as > $ > L^1(\nu) = L^1(\nu_r) \cap L^1(\nu_i) > $ > [!theorem] Radon-Nikodym (Complex) > > Let $\nu$ be a complex measure and $\mu$ be a $\sigma$-finite positive measure, then there exists a complex measure $\lambda$ and an $f \in L^1(\mu)$ such that $\lambda \perp \mu$ and $d\nu = d\lambda + fd\mu$. > > If $\lambda' \perp \mu$ and $d\nu = d\lambda' + f'd\mu$, then $\lambda = \lambda'$ and $f = f'$ $\mu$-a.e. > > *Proof*. By the [[Lebesgue-Radon-Nikodym Theorem]], there are decompositions > $ > d\nu_r = d\lambda_r + f_rd\mu \quad d\nu_i = d\lambda_i + f_id\mu > $ > where $\lambda_r, \lambda_i \perp \mu$ and $f_r, f_i \in L^1(\mu)$. Let $\lambda = \lambda_r + i\lambda_i$ and $f = f_r + if_i$, then $\lambda \perp \mu$ and $f \in L^1(\mu)$, and $d\nu = d\lambda + fd\mu$. > > Let $d\nu = d\lambda' + f'd\mu$ be an alternate decomposition where $\lambda' \perp \mu$, then since $\lambda - \lambda' \perp \mu$, > $ > d\lambda - d\lambda' = (f - f')d\mu > $ > we have an equality of two mutually singular measures, meaning that they are both $0$. Therefore $\lambda = \lambda'$ and $f = f'$ $\mu$-a.e. > [!definition] > > Let $(X, \cm)$ be a measurable space, $\nu$ be a complex measure, and $\mu$ be a positive measure. If $d\nu = fd\mu$, then the **total variation** $\abs{\nu}$ of $\nu$ is the measure $\abs{f}d\mu$. > > *Proof*. Firstly, $\mu = \abs{\nu_r} + \abs{\nu_i}$ is a positive finite measure with $\nu \ll \mu$. By the Radon-Nikodym theorem, there exists $f \in L^1(\mu)$ such that $\nu = d\mu$. > > Let $\mu'$ be another positive measure and $f' \in L^1(\mu')$ such that $d\nu = fd\mu'$. Take $\rho = \mu + \mu'$, then > $ > f \frac{d\mu}{d\rho}d\rho = d\nu = f' \frac{d\mu'}{d\rho}d\rho > $ > and $f\frac{d\mu}{d\rho} = f'\frac{d\mu'}{d\rho}$ $\rho$-a.e. Since $\frac{d\mu}{d\rho}$ and $\frac{d\mu'}{d\rho}$ are both positive, > $ > \abs{f} \frac{d\mu}{d\rho} = \abs{f\frac{d\mu}{d\rho}} = \abs{f'\frac{d\mu'}{d\rho}} = \abs{f'}\frac{d\mu'}{d\rho} \quad \rho\text{-a.e.} > $ > collapsing the equality yields > $ > \abs{f}d\mu = \abs{f}\frac{d\mu}{d\rho}d\rho = \abs{f'}\frac{d\mu'}{d\rho}d\rho = \abs{f'}d\mu' > $ > [!theorem] > > Let $\nu$ be a complex measure on $(X, \cm)$, then > 1. $\abs{\nu(E)} \le \abs{\nu}(E)$ for all $E \in \cm$. > 2. $\nu \ll \abs{\nu}$ and $\abs{\frac{d\nu}{d\abs{\nu}}} = 1$ $\abs{\nu}$-a.e. > 3. $L^1(\nu) = L^1(\abs{\nu})$, and if $f \in L^1(\nu)$, then $\abs{\int f d\nu} \le \int \abs{f} d\abs{\nu}$. > > *Proof.* Let $\mu$ be a positive measure and $f \in L^1(\mu)$ such that $d\nu = fd\mu$, then > $ > \abs{\nu(E)} = \abs{\int_E fd\mu} \le \int_E \abs{f}d\mu = \abs{\nu}(E) > $ > Since $\abs{\nu(E)} \le \abs{\nu}(E)$ for all $E \in \cm$, any $\abs{\nu}$-null set is also a $\nu$-null set. > > Since > $ > \int \abs{g}d\abs{\nu} = \int \abs{g} \cdot \abs{f}d\mu = \int\abs{gf}d\mu < \infty > $ > we have $g \in L^1(\nu)$ if and only if $g \in L^1(\abs{\nu})$.