> [!theorem] > > Let $(X, \cm)$ be a [[Measure Space|measurable space]], $\nu$ be a $\sigma$-finite [[Signed Measure|signed measure]] on $(X, \cm)$, and $\mu$ be a $\sigma$-finite positive measure on $(X, \cm)$. There exist a unique $\sigma$-finite signed measures $\lambda, \rho$ on $(X, \cm)$ such that > $ > \lambda \perp \mu \quad \rho \ll \mu \quad \nu = \lambda + \rho > $ > where $\lambda$ and $\mu$ are [[Mutual Singularity|mutually singular]], and the part $\rho$ is [[Absolute Continuity|absolutely continuous]] with respect to $\mu$. This decomposition $\nu = \lambda + \rho$ is known as the **Lebesgue decomposition** of $\nu$ with respect to $\mu$. > > Moreover, there exists an extended $\mu$-[[Integrable Function|integrable function]] $f: X \to \real$ such that $d\rho = f d\mu$. Any two such functions are equal $\mu$-[[Almost Everywhere|a.e.]] > > If $\nu$ is [[Absolute Continuity|absolutely continuous]] with respect to $\mu$, then $d\nu = fd\mu$ and the [[Equivalence Class|equivalence class]] of all such functions is the **Radon-Nikodym derivative** of $\nu$ with respect to $\mu$, denoted as $d\nu/d\mu$, where > $ > d\nu = \frac{d\nu}{d\mu}d\mu > $ > > *Proof.* > > ### Constructing the Derivative > > First suppose that $\nu$ and $\mu$ are both finite and positive. Let > $ > \cf = \bracs{f: X \to [0, \infty]: \int_E f d\mu \le \nu(E)\ \forall E \in \cm} > $ > be the collection of [[Integral|integrals]] with respect to $\nu$ that try to approximate $\nu$ from below. Note that the approximations from these functions are restricted by the strictly positive sets of $\mu$. If $\nu(E) > 0$ but $\mu(E) = 0$, then no function can satisfy $\int_E fd\mu = \nu(E)$, and no integrals can approximate $\nu$ at this set. > > Since $0 \in \cf$, $\cf$ is non-empty. Let $f, g \in \cf$, and $A = \bracs{x \in X: f(x) > g(x)}$ > $ > \begin{align*} > \int_E\max(f, g)d\mu &= \int_{E \cap A}fd\mu + \int_{E \cap A^c}gd\mu \\ > &\le \nu(E \cap A) + \nu(E \cap A^c)\\ > &= \nu(E) > \end{align*} > $ > and $\max(f, g) \in \cf$. Let $s = \sup\bracs{\int f d\mu: f \in \cf} \le \nu(X)$, and let $\seq{f_n} \subset \cf$ be a [[Sequence|sequence]] such that $\int f_n d\mu \upto s$. Take $f = \sup_{n \in \nat}f_n$, $g_n = \max_{k \le n}f_k$, then $g_n \in \cf$ and $g_n \upto f$ [[Pointwise Convergence|pointwise]]. For any $E \in \cm$, > $ > \int_E g_nd\mu \le \nu(E) \Rightarrow \int_{E}\limv{n}g_nd\mu = > \limv{n}\int_{E} g_nd\mu = \int_E f \le \nu(E) > $ > by the [[Monotone Convergence Theorem]], meaning that $f \in \cf$. Since $\int f d\mu < \infty$, $f < \infty$ $\mu$-a.e. and we can assume that $f < \infty$. > > > ### Decomposition > > Let $d\lambda = d\nu - fd\mu$, then since $f \in \cf$, $d\lambda$ is positive. Moreover, $\lambda$ is singular with respect to $\mu$. Suppose not, then there exists $E \in \cm$ and $\varepsilon > 0$ such that $\mu(E) > 0$ and $E$ is a positive set for $\lambda - \varepsilon \mu$. In that case, let $g = f + \varepsilon \chi_{E}$ and $d\kappa = d\lambda - \varepsilon\chi_E$, then > $ > \kappa(A) = \lambda(A) - \int_{A}\varepsilon\chi_{E}d\mu = \lambda(A) - \varepsilon\mu(E \cap A) \ge 0 > $ > meaning that > $ > d\kappa + gd\mu = d\nu \quad 0 \le \int_E g d\mu \le \nu(E) > $ > so $g \in \cf$ and > $ > \int g d\mu = \int f d\mu + \int \varepsilon\chi_{E}d\mu = \int f d\mu + \varepsilon\mu(E) > $ > Since $\mu(E) > 0$, $\int g d\mu > s$, which contradicts with the fact that it is the supremum. Therefore $d\lambda \perp \mu$. > > > ### Sigma Finiteness > > Now suppose that $\mu, \nu$ are $\sigma$-finite and positive. Then > $ > X = \bigcup_{n \in \nat}U_n = \bigcup_{n \in \nat}V_n \quad \mu(U_n) < \infty, \nu(V_n) < \infty \quad \forall n \in \nat > $ > Assuming that the unions are disjoint, we refine the partitions by taking their intersection. Let $A_n = U_i \cap V_j$ such that > $ > \bigcup_{n \in \nat}A_n = \bigcup_{i \in \nat}\bigcup_{j \in \nat}U_i \cap V_j \quad \mu(A_n) < \infty, \nu(A_n) < \infty > $ > Then we have a sequence of sets $\seq{A_n}$ making up $X$ where each $A_n$ is finite both $\mu$ and $\nu$. > > Define $\mu_n(E) = \mu(E \cap A_n)$ and $\nu_n(E) = \nu(E \cap A_n)$. Then each $\mu_n$ and $\nu_n$ are finite and positive, and for any [[Indicator Function|indicator function]], > $ > \int \chi_E d\mu_n = \mu_n(E) = \mu(E \cap A_n) = \int_{A_n}\chi_{E}d\mu > $ > which extends to all [[Simple Function|simple functions]] by linearity. Then for any $f \in L^+(\mu)$, choose $\seq{\phi_n}$ such that $\phi_n \upto f$ [[Pointwise Convergence|pointwise]] and > $ > \int f d\mu_n =\limv{j}\int\phi_jd\mu_n = \limv{j}\int_{A_n}\phi_jd\mu = \int_{A_n} f d\mu > $ > by the [[Monotone Convergence Theorem]]. Moreover, there is a decomposition from the finite case: > $ > d\nu_n = d\lambda_n + f_nd\mu_n \quad \lambda_n \perp \mu_n > $ > Since $\mu_n(A_n^c) = 0$, we have > $ > \int_{E}f_nd\mu_n = \int_{E \cap A_n}f_nd\mu > $ > meaning that the value of $f_n$ outside of $A_n$ does not matter. This way we can set $f_n(x) = 0 \forall x \not\in A_n$ which allows > $ > \int_E f_n d\mu_n = \int_{E \cap A_n}f_nd\mu = \int_E f_nd\mu > $ > Now, since each $\nu_n$ is null on $A_n^c$, $\lambda_n$ is also null on $A_n^c$. > $ > \nu_n(A_n^c) = \lambda_n(A_n^c) + \int_{A_n^c}f_nd\mu_n \Rightarrow \lambda_n(A_n^c) = 0 > $ > Since $\lambda_n \perp \mu_n$, there exists a partition $X = U \cup V$ such that $U$ is $\lambda_n$-null and $V$ is $\mu_n$-null. Let $U' = U \cup A_n^c$ and $V' = V \cap A_n$. Then $U'$ is $\lambda_n$-null, and > $ > \mu(V') = \mu(V \cap A_n) = \mu_n(V) = 0 > $ > is $\mu$-null, giving us $\lambda_n \perp \mu$. Adding everything up and using the [[Monotone Convergence Theorem]], > $ > \begin{align*} > \nu(E) &= \sum_{n \in \nat}\nu(E \cap A_n) = \sum_{n \in \nat}\nu_n(E) = \sum_{n \in \nat}\nu_n(E \cap A_n) \\ > &= \sum_{n \in \nat}\braks{\lambda_n(E \cap A_n) + \int_{E \cap A_n}f_nd\mu_n} \\ > &= \sum_{n \in \nat}\braks{\lambda_n(E) + \int_{E}f_nd\mu} \\ > &= \underbrace{\sum_{n \in \nat}\lambda_n(E)}_{\lambda} + \int_{E}\underbrace{\sum_{n \in \nat}f_n}_{f} d\mu > \end{align*} > $ > where $\lambda \perp \mu$ and $d\lambda$, $fd\mu$ are $\sigma$-finite. Since each part $f_n \in L^1(\mu)$, $f_n < \infty$ $\mu$-a.e., we can replace the infinity value with any finite value, and assume that $f < \infty$. > > > ### Signed Measure > > Now suppose that $\mu$ is a positive $\sigma$-finite measure and $\nu$ is a signed measure. Let $\nu = \nu^+ - \nu^-$ be its [[Jordan Decomposition Theorem|Jordan decomposition]], then we can decompose each piece as > $ > d\nu^+ = d\lambda^+ + f^+d\mu \quad d\nu^- = d\lambda^- + f^-d\mu > $ > Adding together the parts yields > $ > d\nu = \underbrace{d\lambda^+ - d\lambda^-}_{\lambda} + \underbrace{(f ^+ - f^{-})}_{f}d\mu > $ > Since one of $\nu^+$ and $\nu^-$ is finite, and $f^+$ and $f^-$ can both be assumed to be finite, the subtraction is defined and the decomposition yields $d\lambda$ and $fd\mu$ both being $\sigma$-finite signed measures. > > > ### Uniqueness > > Let $d\nu = d\lambda' + f'd\mu$ be another decomposition where $f': X \to [0, \infty)$ and $\lambda' \perp \mu$. Then > $ > d\lambda - d\lambda' = f'd\mu - fd\mu = (f' - f) d\mu > $ > Since $\lambda, \lambda' \perp \mu$, $\lambda - \lambda' \perp \mu$ as well. As $f' - f$ is still integrable, $(f' - f)d\mu \ll \mu$, and $\lambda - \lambda' \perp (f' - f)d\mu$. Let $X = U \cup V$ be the decomposition such that $U$ is $(\lambda - \lambda')$-null and $V$ is $(f' - f)d\mu$-null. Then > $ > \begin{align*} > (\lambda - \lambda')(E) &= 0 &\forall E \in \cm, E \subseteq U \\ > \int_{E}(f' - f)d\mu &= 0 &\forall E \in \cm, E \subseteq V > \end{align*} > $ > Therefore for any $E \in \cm$, > $ > \begin{align*} > (\lambda - \lambda')(E) &= (\lambda - \lambda')(E \cap U) + (\lambda - \lambda')(E \cap V) = 0 \\ > \int_{E}(f' - f)d\mu &= \int_{E \cap U}(f' - f)d\mu + \int_{E \cap V}(f' - f)d\mu = 0 > \end{align*} > $ > and the decomposition $d\nu = d\lambda + d\rho$ where $d\rho = fd\mu$ is unique. Since $\int_E fd\mu = \int_E f'd\mu$ for all $E \in \cm$, $f = f'$ $\mu$-a.e.