From 98127388ec0a67d2334285e1a0f0c1cf58b666cf Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Tue, 30 Jun 2026 13:42:43 -0400 Subject: [PATCH] Retracted localisable version of Radon-Nikodym. --- src/measure/vector/rn.tex | 100 -------------------------------------- 1 file changed, 100 deletions(-) diff --git a/src/measure/vector/rn.tex b/src/measure/vector/rn.tex index 7cbd27a..4bb0d8a 100644 --- a/src/measure/vector/rn.tex +++ b/src/measure/vector/rn.tex @@ -97,105 +97,5 @@ \end{proof} -\begin{theorem}[Lebesgue-Radon-Nikodym (Localisable)] -\label{theorem:lebesgue-radon-nikodym-localisable} - Let $(X, \cm)$ be a measurable space, $\mu, \nu: \cm \to [0, \infty]$ be positive measures, and $\cf \subset \cm$ be an ideal such that: - \begin{enumerate}[label=(\alph*)] - \item $(X, \cm, \mu + \nu)$ and $(X, \cm, \mu)$ are localisable. - \item $\cf$ is a \hyperref[scaffold]{definition:measure-scaffold} for both $\mu$ and $\nu$. - \end{enumerate} - - then there exists a unique pair of positive measures $\nu_a, \nu_s: \cm \to [0, \infty]$ such that: - \begin{enumerate} - \item $\nu = \nu_a + \nu_s$. - \item $\nu_a$ is absolutely continuous with respect to $\mu$. - \item $\nu_s$ is mutually singular with $\mu$. - \item $\cf$ is a scaffold for $\nu_a$ and $\nu_s$. - \end{enumerate} - - The pair $(\nu_a, \nu_s)$ is the \textbf{Lebesgue decomposition} of $\nu$ with respect to $\mu$. Moreover, there exists a unique $\frac{d\nu_a}{d\mu} \in L^+(X, \cm, \mu)$ such that for every $A \in \cm$, - \[ - \nu_a(A) = \int_{A} \frac{d\nu_a}{d\mu} d\mu - \] - - If $\nu \ll \mu$, then $\nu_s = 0$ and $\nu(dx) = \frac{d\nu_a}{d\mu}\mu(dx) = \frac{d\nu_a}{d\mu}\mu(dx)$. In which case, the function $\frac{d\nu}{d\mu}$ is the \textbf{Radon-Nikodym derivative} of $\nu$ with respect to $\mu$. -\end{theorem} -\begin{proof} - For each $A \in \cf$ and $E \in \cm$, let - \[ - \mu^A(E) = \mu(A \cap E) \quad \nu^A(E) = \nu(A \cap E) - \] - - then $\mu^A$ and $\nu^A$ are finite measures on $A$. By the \hyperref[finite case]{theorem:lebesgue-radon-nikodym}, there exists an a.e. unique $f_A \in L^+(A, \mu)$ and $\nu_s^A: \cm \to [0, \infty]$ such that: - \begin{enumerate} - \item $d\nu^A = f_A\mu^A + \nu_s^A = f_A\mu + \nu_s^A$. - \item $\nu_s^A$ is mutually singular with $\mu$. - \end{enumerate} - - The uniqueness given by the finite case implies that for any $A, B \in \cf$, $f_A|_{A \cap B} = f_B|_{A \cap B}$ $\mu$-almost everywhere. By the \hyperref[gluing lemma for measurable functions]{lemma:gluing-measurable}, there exists $f \in L^+(X, \mu)$ such that $f|_A = f_A$ $\mu$-almost everywhere for all $A \in \cf$. - - By the \hyperref[gluing lemma for measures]{lemma:gluing-measure}, - \[ - \nu_s: \cm \to [0, \infty] \quad E \mapsto \sup_{A \in \cf}\nu_s^A(E \cap A) - \] - - is a measure. - - (4): By definition, $\cf$ is a scaffold for $\nu_s$. By \autoref{lemma:scaffolded-ac}, it is also a scaffold for $f d\mu$. - - (1): Let $E \in \cm$, then since $\cf$ is a scaffold for $\nu$, - \begin{align*} - \nu(E) &= \sup_{A \in \cf}\nu(A \cap E) = \sup_{A \in \cf}\nu^A(A \cap E) \\ - &= \sup_{A \in \cf}\braks{\int_{A \cap E}f_A d\mu + \nu_s^A(A \cap E)} - \end{align*} - - As $\cf$ is an ideal, for any $A, B \in \cf$, $A \cup B \in \cf$, and - \begin{align*} - \int_{A \cap E}f_A d\mu \vee \int_{B \cap E}f_B d\mu &\le \int_{(A \cup B) \cap E}f_{A \cup B} d\mu \\ - \nu_s^A(A \cap E) \vee \nu_s^B(B \cap E) &\le \nu_s^{A \cup B}((A \cup B) \cap E) - \end{align*} - - Thus the sum and the supremum may be interchanged, so - \[ - \nu(E) = \sup_{A \in \cf}\int_{A \cap E}f_A d\mu + \sup_{A \in \cf}\nu_s^A(A \cap E) - \] - - Now, since $\cf$ is a scaffold for $f\mu$, - \[ - \sup_{A \in \cf}\int_{A \cap E}f_A d\mu = \int_A f d\mu - \] - - By definition of $\nu_s$, - \[ - \sup_{A \in \cf}\nu_s^A(A \cap E) = \nu_s(E) - \] - - Therefore $\nu(E) = \int_E f d\mu + \nu_s(E)$, so $\nu(dx) = f(x)\mu(dx) + \nu_s(dx)$. - - (2): $\nu_a(dx) = f(x)\mu(dx)$. - - (3): For each $A \in \cf$, $f_A d\mu \perp \nu_s^A$, so there exists $E_A, F_A \in \cm$ such that: - \begin{enumerate}[label=(\roman*)] - \item $A = E_A \sqcup F_A$. - \item $\mu(F_A) = 0$ and $\nu_s^A(E_A) = 0$. - \end{enumerate} - - Let $F$ be an essential supremum of $\bracs{F_A}_{A \in \cf}$ with respect to $(\mu + \nu)$. By \autoref{lemma:gluing-measurable-sets}, for each $A \in \cf$, - \[ - \mu(F \cap A) = \mu(F \cap E_A) \le (\mu + \nu)(F \cap E_A) = 0 - \] - - Since $\cf$ is a scaffold for $\mu$, $\mu(F) = 0$. On the other hand, for each $A \in \cf$, - \begin{align*} - \nu_s(A \setminus F) &= \nu_s(E_A \setminus F) + \nu_s(F_A \setminus F) \\ - &\le \nu_s(E_A) + (\mu + \nu)(F_A \setminus F) = 0 - \end{align*} - - By definition of $\nu_s$, $\nu(F^c) = 0$. Therefore $\mu \perp \nu_s$. - - (Uniqueness): By uniqueness of the finite case, the restriction of $(\nu_a, \nu_s)$ to each set in $\cf$ is unique. Since $\cf$ is a scaffold for both measures, they are uniquely determined by their restrictions to each set in $\cf$. - -\end{proof} -