diff --git a/src/measure/index.tex b/src/measure/index.tex
index 321fb91..1b28cd8 100644
--- a/src/measure/index.tex
+++ b/src/measure/index.tex
@@ -8,4 +8,5 @@
 \input{./measurable-maps/index.tex}
 \input{./lebesgue-integral/index.tex}
 \input{./bochner-integral/index.tex}
+\input{./differentiation/index.tex}
 \input{./notation.tex}
diff --git a/src/measure/measure/measure.tex b/src/measure/measure/measure.tex
index 1c07b16..6fcd13c 100644
--- a/src/measure/measure/measure.tex
+++ b/src/measure/measure/measure.tex
@@ -36,7 +36,7 @@
     Let $(X, \cm, \mu)$ be a measure space. A statement holds \textbf{$\mu$-almost everywhere/$\mu$-a.e./a.e.} if it is false on a subset of a $\mu$-null set.
 \end{definition}
 
-\begin{proposition}[{{\cite[Theorem 1.8]{Folland}}}]
+\begin{proposition}
 \label{proposition:measure-properties}
     Let $(X, \cm, \mu)$ be a measure space, then:
     \begin{enumerate}
@@ -56,7 +56,7 @@
 
     \end{enumerate}
 \end{proposition}
-\begin{proof}
+\begin{proof}[Proof, {{\cite[Theorem 1.8]{Folland}}}. ]
     (1): $\mu(F) = \mu(E) + \mu(F \setminus E) \ge \mu(E)$.
 
     (2): For each $n \in \natp$, let $F_n = E_n \setminus \bigcup_{k = 1}^{n-1}E_k$, then $\seq{F_n}$ is pairwise disjoint with $\bigsqcup_{n \in \natp}F_n = \bigcup_{n \in \natp}E_n$. In which case,
diff --git a/src/measure/vector/ac.tex b/src/measure/vector/ac.tex
index b779f39..f03bfcf 100644
--- a/src/measure/vector/ac.tex
+++ b/src/measure/vector/ac.tex
@@ -1,10 +1,37 @@
-\section{Absolutely Continuous}
-\label{section:absolutely-continuous-measure}
+\section{Absolute Continuity}
+\label{section:absolute-continuity}
 
 \begin{definition}[Absolutely Continuous]
 \label{definition:absolutely-continuous}
-    Let $(X, \cm)$ be a measurable space and $\mu, \nu$ be signed/vector measures on $X$, then $\nu$ is \textbf{absolutely continuous} with respect to $\mu$, denoted $\nu \ll \mu$, if every $\mu$-null set is $\nu$-null.
+    Let $(X, \cm)$ be a measurable space, $\mu$ be a positive measure on $(X, \cm)$, and $\nu$ be a signed measure or vector veasure on $(X, \cm)$, then $\nu$ is \textbf{absolutely continuous} with respect to $\mu$, denoted $\nu \ll \mu$, if for every $E \in \cm$ with $\mu(E) = 0$, $\nu(E) = 0$ as well. 
 \end{definition}
 
+\begin{lemma}
+\label{lemma:ac-gymnastics}
+    Let $(X, \cm)$ be a measurable space, $\mu$ be a positive measure on $(X, \cm)$, and $\nu$ be a signed measure or vector measure on $(X, \cm)$, then $\nu \ll \mu$ if and only if $|\nu| \ll \mu$. 
+\end{lemma}
 
+\begin{proposition}
+\label{proposition:ac-epsilon-delta}
+    Let $(X, \cm)$ be a measurable space, $\mu$ be a positive measure on $(X, \cm)$, $E$ be a normed space over $K \in \RC$, and $\nu: \cm \to E$ be a vector measure, then the following are equivalent:
+    \begin{enumerate}
+        \item $\nu$ is absolutely continuous with respect to $\mu$. 
+        \item For each $\eps > 0$, there exists $\delta > 0$ such that for any $A \in \cm$ with $\mu(A) < \delta$, $\norm{\nu(A)}_E < \eps$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Theorem 3.15]{Folland}}}. ]
+    $(\neg 2) \Rightarrow (\neg 1)$: Using \autoref{lemma:ac-gymnastics}, assume without loss of generality that $\nu$ is a positive finite measure. 
+
+    By assumption, there exists $\eps > 0$ and $\seq{A_n} \subset \cm$ such that for each $n \in \natp$, $\mu(A_n) < 2^{-n}$ and $\norm{\nu(A_n)}_E \ge \eps$. For each $N \in \natp$, let $B_N = \bigcup_{n \ge N}A_n$, then $\mu(B_N) < 2^{-N+1}$. By \autoref{proposition:measure-properties},
+    \[
+    \mu\paren{\bigcap_{N \in \natp}B_N} = \limv{N}\mu(B_N) = 0
+    \]
+
+    but by \autoref{proposition:vector-measure-properties}, 
+    \[
+    \norm{\nu\paren{\bigcap_{N \in \natp}B_N}}_E = \limv{N}\norm{\nu(B_N)}_E \ge \eps
+    \]
+
+    so (1) does not hold. 
+\end{proof}
 
diff --git a/src/measure/vector/index.tex b/src/measure/vector/index.tex
index d0dd4dc..50bce95 100644
--- a/src/measure/vector/index.tex
+++ b/src/measure/vector/index.tex
@@ -3,7 +3,8 @@
 
 \input{./complex.tex}
 \input{./vector.tex}
+\input{./variation.tex}
 \input{./ac.tex}
 \input{./ms.tex}
-\input{./variation.tex}
+\input{./ms.tex}
 
diff --git a/src/measure/vector/rn.tex b/src/measure/vector/rn.tex
new file mode 100644
index 0000000..d274633
--- /dev/null
+++ b/src/measure/vector/rn.tex
@@ -0,0 +1,98 @@
+\section{The Lebesgue-Radon-Nikodym Theorem}
+\label{section:lebesgue-radon-nikodym}
+
+\begin{theorem}[Lebesgue-Radon-Nikodym]
+\label{theorem:lebesgue-radon-nikodym}
+    Let $(X, \cm)$ be a measurable space, $\mu$ be a $\sigma$-finite positive measure on $(X, \cm)$, $H$ be a Hilbert space over $K \in \RC$ and $\nu: \cm \to H$ be a finite vector measure, then there exists a unique pair of finite vector measures $\nu_a, \nu_s: \cm \to H$ 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 $\mu$. 
+    \end{enumerate}
+
+    The pair $(\nu_a, \nu_s)$ is the \textbf{Lebesgue decomposition} of $\nu$ with respect to $\mu$. Moreover, there exists a unique $\frac{d\nu_a}{d\mu} \in L^1(X, \cm, \mu; H)$ 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 $\nu(dx) = \frac{d\nu}{d\mu}\mu(dx)$. 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}[Proof, {{\cite[Exercise 6.18]{Folland}}}\footnote{I decided to abuse Hilbert spaces for this theorem because it is more fun, and because I will use the Riesz representation theorem twice.}. ]
+    (Finite + Positive): First suppose that $\mu$ is finite and $\nu$ is positive. Let $\lambda = \mu + \nu$, then the mapping
+    \[
+    I_\nu: L^2(X, \cm, \lambda; K) \to K \quad f \mapsto \int f d\nu
+    \]
+
+    is a continuous linear functional on $L^2(X, \cm, \lambda; K)$. By the \hyperref[Riesz Representation Theorem]{theorem:riesz-hilbert}, there exists a unique $g \in L^2(X, \cm, \lambda; K)$ such that for every $f \in L^2(X, \cm, \mu; K)$, $\int f d\nu = \int f g d\lambda$. By uniqueness of the Riesz representation, $\int f d\mu = \int f(1 - g)d\lambda$ for all $f \in L^2(X, \cm, \lambda; K)$. 
+
+    Since $0 \le \int f d\nu \le \norm{f}_{L^1(X, \cm, \lambda; K)}$ for all $f \in L^2(X, \cm, \lambda; K) \cap L^+(X, \cm)$, $0 \le g \le 1$ $\lambda$-almost everywhere. Let $A = \bracs{g < 1}$, $B = \bracs{g = 1}$, 
+    \[
+    \nu_a: \cm \to H \quad E \mapsto \nu(E \cap A)
+    \]
+    
+    and
+    \[
+    \nu_s: \cm \to H \quad E \mapsto \nu(E \cap B)
+    \]
+
+    then
+    \begin{enumerate}
+        \item By definition, $\nu = \nu_a + \nu_s$. 
+        \item Let $E \in \cm$ with $\mu(E) = 0$, then $(1 - g)|_E = 0$ $\lambda$-almost everywhere. Thus $E \subset B$ modulo a $\lambda$-null set, and
+        \[
+            \nu_a(E) = \nu(E \cap A) \le \lambda(A \cap B) = 0
+        \]
+
+        so $\nu_a \ll \mu$. 
+        
+        \item Since $\mu(B) = 0$ and $\nu_s(A) = 0$, $\mu \perp \nu_s$. 
+    \end{enumerate}
+    
+    Let
+    \[
+    h: X \to [0, \infty) \quad x \mapsto \frac{\one_A(x)g(x)}{1 - g(x)}
+    \]
+
+    then for each $E \in \cm$, 
+    \begin{align*}
+        \nu_a(E) &= \int_{E \cap A} g d\lambda = \int_E \one_A g d\lambda \\
+        &= \int_{E \cap A} \frac{\one_A g}{1 - g} \cdot (1 - g)d\lambda = \int_E \frac{\one_A g}{1 - g}d\mu
+    \end{align*}
+    
+    so $\one_A g/(1 - g)$ is the desired Radon-Nikodym derivative. 
+
+    ($\sigma$-Finite + Positive): Now suppose that $\mu$ is $\sigma$-finite and $\nu$ is still positive. Let $\seq{E_n} \subset \cm$ be a pairwise disjoint sequence such that $\bigsqcup_{n \in \natp}E_n = X$ and $\mu(E_n) < \infty$ for all $n \in \natp$. For each $n \in \natp$ and $E \in \cm$, let $\mu_n(E) = \mu(E \cap E_n)$ and $\nu_n(E) = \nu(E \cap E_n)$, then there exists $g_n \in L^1(X, \cm, \mu)$ and a positive measure $\nu_s^{(n)}$ such that
+    \begin{enumerate}
+        \item $\nu_n(dx) = g_n\mu(dx) + \nu_s^{(n)}(dx)$. 
+        \item[(3)] $\nu_s^{(n)}$ is mutually singular with $\mu$. 
+    \end{enumerate}
+
+    Let $g = \sum_{n \in \natp}g_n$ and $\nu_s = \sum_{n \in \natp}\nu_s^{(n)}$, then by the \hyperref[Monotone Convergence Theorem]{theorem:mct}, completeness of $L^1(X, \cm, \mu)$, and \hyperref[completeness of $M(X, \cm; \real)$]{definition:vector-measure-finite-space}, $g \in L^1(X, \cm, \mu)$ and $\nu_s$ is a measure with
+    \begin{enumerate}
+        \item $\nu(dx) = g\mu(dx) + \nu_s$. 
+        \item[(3)] $\nu_s$ is mutually singular with $\mu$. 
+    \end{enumerate}
+
+    (Hilbert): Finally, suppose that $\nu$ is an $H$-valued finite vector measure, then the mapping
+    \[
+    I_\nu: L^2(X, \cm, |\nu|; H) \to K \quad f \mapsto \int f d\nu
+    \]
+
+    is a continuous linear functional. By the \hyperref[Riesz Representation Theorem]{theorem:riesz-hilbert}, there exists a unique $g \in L^2(X, \cm, |\nu|; H)$ such that for every $f \in L^2(X, \cm, |\nu|; H)$, $\int f d\nu = \int fg d|\nu|$. For each $E \in \cm$, let $\nu_a(E) = \int_E g d|\nu|_a$ and $\nu_s(E) = \int_E g d|\nu|_s$, then
+    \begin{enumerate}
+        \item $\nu(E) = \int_E g d|\nu| = \int_E g d|\nu|_a + \int_E g d|\nu|_s = \nu_a(E) + \nu_s(E)$. 
+        \item Since $|\nu|_a$ is absolutely continuous with respect to $\mu$, so is $\nu_a$. 
+        \item Since $|\nu|_s$ is mutually singular with $\mu$, so is $\nu_s$. 
+    \end{enumerate}
+
+    and
+    \[
+    \nu_a(E) = \int_E g d|\nu|_a = \int_E g \cdot \frac{d|\nu|_a}{d\mu} d\mu
+    \]
+
+    so $\frac{d\nu_a}{d\mu} = g \cdot \frac{d|\nu|_a}{d\mu}$ is the desired Radon-Nikodym derivative. 
+    
+    
+    (Uniqueness): For any decomposition $\nu = \rho_a + \rho_s$ satisfying the above, then $\rho_a - \nu_a = \nu_s - \rho_s$ with $\rho_a - \nu_a \perp \nu_s - \rho_s$. Therefore $\rho_a = \nu_a$, $\rho_s = \nu_s$, and the decomposition is unique. 
+\end{proof} 
+
diff --git a/src/measure/vector/vector.tex b/src/measure/vector/vector.tex
index 5f7f929..f120833 100644
--- a/src/measure/vector/vector.tex
+++ b/src/measure/vector/vector.tex
@@ -42,5 +42,26 @@
 
 \end{proof}
 
+\begin{proposition}
+\label{proposition:vector-measure-properties}
+    Let $(X, \cm)$ be a measurable, $E$ be a normed vector space over $K \in \RC$, and $\mu: \cm \to E$ be a vector measurethen:
+    \begin{enumerate}
+        \item For any $\seq{A_n} \subset \cm$ with $A_n \subset A_{n+1}$ for all $n \in \nat$, $\mu\paren{\bigcup_{n \in \nat}A_n} = \limv{n}\mu(A_n)$.
+        \item For any $\seq{A_n} \subset \cm$ with $A_n \supset A_{n+1}$ for all $n \in \nat$, $\mu(\bigcap_{n \in \natp}A_n) = \limv{n}\mu(A_n)$.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): Let $A_0 = \emptyset$. For each $n \in \natp$, let $B_n = A_n \setminus B_{n-1}$, then $\bigcup_{n \in \natp}A_n = \bigsqcup_{n \in \natp}B_n$ and
+    \[
+    \limv{n}\mu(A_n) = \sum_{n = 1}^\infty \mu(B_n) = \mu\paren{\bigcup_{n \in \natp}A_n}
+    \]
+
+    (2): By (1), 
+    \begin{align*}
+        \limv{n}\mu(A_n) &= \mu(A_1) - \limv{n}\mu(A_1 \setminus A_n) \\
+        &= \mu(A_1) - \mu\paren{A_1 \setminus \bigcup_{n \in \natp}A_n} = \mu\paren{\bigcap_{n \in \natp}A_n}
+    \end{align*}
+     
+\end{proof}