Swapped in vocabulary.

src/fa/lp/index.tex

@@ -6,3 +6,4 @@
 \input{./duality.tex}
 \input{./ui.tex}
 \input{./seq.tex}
+\input{./local.tex}

src/fa/lp/local.tex

@@ -0,0 +1,44 @@
+\section{Locally Integrable Functions}
+\label{section:locally-integrable}
+
+\begin{definition}[Locally Integrable*]
+\label{definition:locally-integrable}
+    Let $(X, \cm, \cf, \mu)$ be a \hyperref[scaffolded]{definition:measure-scaffold} measure space, $E$ be a normed vector space over $K \in \RC$, $f: X \to E$ be strongly measurable, and $p \in [1, \infty]$, then $f$ is \textbf{locally $p$-integrable} if for every $A \in \cf$, $\int_A \norm{f}_A^p d\mu < \infty$. 
+
+    The set $\mathcal{L}^p_\cf(X, \cm, \mu; E) = \mathcal{L}^p_\cf(X; E) = \mathcal{L}^p_\cf(\mu; E)$ is the space of all locally $p$-integrable $E$-valued functions on $X$. 
+\end{definition}
+
+\begin{definition}[Locally Bounded*]
+\label{definition:locally-bounded-measure}
+    Let $(X, \cm, \cf, \mu)$ be a \hyperref[scaffolded]{definition:measure-scaffold} measure space, $E$ be a normed vectorr space over $K \in \RC$, and $f: X \to E$ be strongly measurable, then $f$ is \textbf{essentially bounded} if for every $A \in \cf$, $\norm{\one_A f}_{L^\infty(A; E)} < \infty$. 
+
+    The set $\mathcal{L}^\infty_\cf(X, \cm, \mu; E) = \mathcal{L}^\infty_\cf(X; E) = \mathcal{L}^\infty_\cf(\mu; E)$ is the space of all locally bounded $E$-valued functions on $X$. 
+\end{definition}
+
+\begin{definition}[Local $L^p$ Space*]
+\label{definition:local-lp-space}
+    Let $(X, \cm, \cf, \mu)$ be a \hyperref[scaffolded]{definition:measure-scaffold} measure space, $E$ be a normed vector space over $K \in \RC$, and $p \in [1, \infty]$. For each $A \in \cf$ and $f \in \mathcal{L}^p_\cf(X, \cm, \mu; E)$, let $[f]_{L^p_A(X; \mu)} = \norm{\one_A f}_{L^p(X; \mu)}$, then the $[\cdot]_{L^p_A(X; \mu)}$ is a seminorm on \hyperref[$\mathcal{L}^p_\cf(X; E)$]{definition:locally-integrable}. The set
+    \[
+    L^p_\cf(X, \cm, \mu; E) = \mathcal{L}^p_\cf(X, \cm, \mu; E)/\bracs{f|f = 0\text{ a.e.}}
+    \]
+
+    equipped with the seminorms $\bracsn{[\cdot]_{L^p_A(X;\mu)}|A \in \cf}$ is a separated locally convex space, and the \textbf{local $L^p$ space} of $(X, \cm, \mu)$. 
+\end{definition}
+
+\begin{lemma}
+\label{lemma:gluing-local-lp}
+    Let $(X, \cm, \cf, \mu)$ be a \hyperref[scaffolded]{definition:measure-scaffold} localisable measure space, $E$ be a normed vector space over $K \in \RC$, $p \in [1, \infty]$, and $\bracsn{f_A}_{A \in \cf}$ such that:
+    \begin{enumerate}[label=(\alph*)]
+        \item For each $A \in \cf$, $f_A \in \mathcal{L}^p(A; E)$. 
+        \item For each $A, B \in \cf$, $f_A|_{A \cap B} = f_B|_{A \cap B}$ almost everywhere.
+        \item $\bigcup_{A \in \cf}f_A(A)$ is a separable subset of $E$. 
+    \end{enumerate}
+
+    then there exists a unique $f \in L^p_\cf(X; E)$ such that $f|_A = f_A$ for all $A \in \cf$. 
+\end{lemma}
+\begin{proof}
+    By the \hyperref[gluing lemma for measurable functions]{lemma:gluing-measurable}.
+\end{proof}
+
+
+

src/measure/vector/fin.tex

@@ -66,7 +66,7 @@ In any case, the above example shows that a linear functional on $M(X, \cm; \com
     
     In other words,
     \begin{enumerate}[start=2]
-        \item There exists a semifinite measure space $(\Omega, \cn, \mu)$ such that $\mathscr{M} \iso L^1(\Omega; H)$ and $\mathscr{M}^* \iso L^\infty(\Omega; H)$. 
+        \item There exists a decomposable measure space $(\Omega, \cn, \mu)$ such that $\mathscr{M} \iso L^1(\Omega; H)$ and $\mathscr{M}^* \iso L^\infty(\Omega; H)$. 
     \end{enumerate}
 \end{theorem}
 \begin{proof}[Proof, {{\cite{StackRadonDual}}}. ]

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}
-