diff --git a/src/fa/lp/definition.tex b/src/fa/lp/definition.tex
index 26d3f4f..8c46c83 100644
--- a/src/fa/lp/definition.tex
+++ b/src/fa/lp/definition.tex
@@ -52,23 +52,22 @@
 
 \begin{theorem}[Hölder's Inequality]
 \label{theorem:holder}
-    Let $(X, \cm, \mu)$ be a measure space, $E, F$ be a normed vector spaces, $p, q \in [1, \infty]$ be Hölder conjugates, then for any $f \in \mathcal{L}^p(X; E)$ and $g \in \mathcal{L}^q(X; F)$,
+    Let $(X, \cm, \mu)$ be a measure space and $p, q \in [1, \infty]$ be Hölder conjugates, then for any $f, g: \in L^+(X)$, 
     \[
-    \int \norm{f}_E \norm{g}_F d\mu \le \norm{f}_{L^p(X; E)}\norm{g}_{L^q(X; F)}
+    \int fg d\mu \le \norm{f}_{L^p(X; \real)}\norm{g}_{L^q(X; \real)}
     \]
 
 \end{theorem}
 \begin{proof}[Proof, {{\cite[Theorem 6.2]{Folland}}}. ]
     First suppose that $p = 1$ and $q = \infty$. In this case,
     \[
-    \int \norm{f}_E \norm{g}_F d\mu \le \norm{g}_{L^\infty(X; F)}\int \norm{f}_Ed\mu = \norm{f}_{L^1(X; E)}\norm{g}_{L^\infty(X; F)}
+    \int fg d\mu \le \norm{g}_{L^\infty(X; \real)}\int fd\mu = \norm{f}_{L^1(X; \real)}\norm{g}_{L^\infty(X; \real)}
     \]
 
-    Now suppose that $p, q \in (1, \infty)$ are Hölder conjugates. Assume without loss of generality that $\norm{f}_{L^p(X; E)} = \norm{g}_{L^q(X; F)} = 1$. By \hyperref[Young's inequality]{lemma:young-inequality},
+    Now suppose that $p, q \in (1, \infty)$ are Hölder conjugates, and assume without loss of generality that $\norm{f}_{L^p(X; \real)} = \norm{g}_{L^q(X; \real)} = 1$. In which case, by \hyperref[Young's inequality]{lemma:young-inequality},
     \[
-    \int \norm{f}_E \norm{g}_F d\mu \le \int \frac{\norm{f}_E^p}{p} + \frac{\norm{g}_F^q}{q} d\mu = \frac{1}{p}\int \norm{f}_E d\mu + \frac{1}{q}\int \norm{g}_F^q d\mu = 1
+    \int fg d\mu \le \int \frac{f^p}{p} + \frac{g^q}{q} d\mu = \frac{1}{p}\int f d\mu + \frac{1}{q}\int g d\mu = 1
     \]
-
 \end{proof}
 
 \begin{theorem}[Minkowski's Inequality, {{\cite[6.5]{Folland}}}]
diff --git a/src/fa/lp/ineq.tex b/src/fa/lp/ineq.tex
index 3f9ab6f..8610907 100644
--- a/src/fa/lp/ineq.tex
+++ b/src/fa/lp/ineq.tex
@@ -48,3 +48,50 @@
     \end{align*}
     
 \end{proof}
+
+
+\begin{theorem}[Minkowski's Inequality for Integrals]
+\label{theorem:minkowski-integral}
+    Let $(X, \cm, \mu)$ and $(Y, \cn, \nu)$ be $\sigma$-finite measure spaces, $E$ be a Banach space, and $f: X \times Y \to E$ be $(\cm \otimes \cn)$-strongly measurable. Then:
+    \begin{enumerate}
+        \item If $E = \real$ and $f \ge 0$, then for any $p \in [1, \infty)$, 
+        \[
+        \braks{\int_{X}\paren{\int_Y f(x, y)\nu(dy)}^{p}\mu(dx)}^{1/p} \le \int_{Y}\norm{f(\cdot, y)}_{L^p(X, \mu; E)} \nu(dy)
+        \]
+        \item For any $p \in [1, \infty]$, if
+        \begin{enumerate}[label=(\alph*)]
+            \item For $\nu$-almost every $y \in Y$, $f(\cdot, y) \in L^p(\mu; E)$.
+            \item The mapping $y \mapsto \norm{f(\cdot, y)}_{L^p(\nu; E)}$ is in $L^1(\mu; \real)$.
+        \end{enumerate}
+
+        then the mapping $x \mapsto \int_Y f(x, y)\nu(dy)$ is in $L^p(\mu; E)$, with
+        \[
+        \norm{\int_Y f(\cdot, y)\nu(dy)}_{L^p(X, \mu; E)} \le \int_{Y}\norm{f(\cdot, y)}_{L^p(X, \mu; E)} \nu(dy)
+        \]
+    \end{enumerate}
+    
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Theorem 6.19]{Folland}}}. ]
+    (1): If $p = 1$, then by \hyperref[Tonelli's Theorem]{theorem:fubini-tonelli}, 
+    \[
+    \int_X\int_Y f(x, y) \nu(dy)\mu(dx) = \int_Y\int_X f(x, y)\mu(dx)\nu(dy)
+    \]
+
+    Now suppose that $p \in (1, \infty)$, assume without loss of generality that $\int_{Y}\norm{f(\cdot, y)}_{L^p(X, \mu; E)} \nu(dy) < \infty$. Let $q$ be the Hölder conjugate of $p$ and $\phi \in L^q(\mu; \real) \cap L^+(\mu)$ with $\phi \ge 0$, then by \hyperref[Tonelli's Theorem]{theorem:fubini-tonelli},
+    \begin{align*}
+        \int_X \int_Y f(x, y)\nu(dy) \phi(x) \mu(dx) &= \int_X\int_Y  f(x, y)\phi(x)\nu(dy) \mu(dx) \\
+        &= \int_Y\int_X f(x, y) \phi(x) \mu(dx) \nu(dy)
+    \end{align*}
+    
+
+    By \hyperref[Hölder's Inequality]{theorem:holder}, 
+    \[
+    \int_Y\int_X f(x, y) \phi(x) \mu(dx) \nu(dy) \le \norm{\phi}_{L^q(\mu; \real)}\int_Y \norm{f(\cdot, y)}_{L^p(\mu; \real)} \nu(dy)
+    \]
+    
+    As $f \ge 0$ and the above holds for all $\phi \in L^q(\mu; \real) \cap L^+(\mu)$, \autoref{theorem:lp-dual-function} implies that $x \mapsto \int_Y f(x, y)\nu(dy) \in L^p(\mu)$ with 
+    \[
+    \braks{\int_{X}\paren{\int_Y f(x, y)\nu(dy)}^{p}\mu(dx)}^{1/p} \le \int_{Y}\norm{f(\cdot, y)}_{L^p(X, \mu; E)} \nu(dy)
+    \]
+\end{proof}
+
diff --git a/src/fa/lp/weak.tex b/src/fa/lp/weak.tex
new file mode 100644
index 0000000..0d68c47
--- /dev/null
+++ b/src/fa/lp/weak.tex
@@ -0,0 +1,45 @@
+\section{Lorentz Spaces}
+\label{section:weak-lp}
+
+\begin{definition}[Distribution Function]
+\label{definition:weak-lp-distribution}
+    Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed vector space, and $f \in L^0(X; E)$ be strongly measurable, then the \textbf{distribution function} of $f$ is the mapping
+    \[
+    \lambda_f: (0, \infty) \to [0, \infty] \quad \alpha \mapsto \mu\bracsn{\norm{f}_E > \alpha}
+    \]
+\end{definition}
+
+\begin{proposition}
+\label{proposition:weak-lp-distribution-properties}
+    Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed space, and $f, g: X \to E$ be strongly measurable, then:
+    \begin{enumerate}
+        \item $\lambda_f$ is non-increasing and right-continuous.
+        \item If $|f| \le |g|$, then $\lambda_f \le \lambda_g$. 
+        \item For each $\alpha > 0$ and $t \in (0, 1)$, $\lambda_{f + g}(\alpha) \le \lambda_f(t\alpha) + \lambda_g((1 - t)\alpha)$. 
+        \item For any strongly measurable functions $\seq{f_n} \subset E^X$, if $\norm{f_n}_E \upto \norm{f}_{E}$ pointwise, then $\lambda_{f_n} \upto \lambda_f$ pointwise.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): Let $\alpha > 0$ and $\seq{\alpha_n} \subset (\alpha, \infty)$ such that $\alpha_n \downto \alpha$, then
+    \[
+        \mu\bracsn{\norm{f}_E > \alpha} = \mu\paren{\bigcup_{n \in \natp}\bracsn{\norm{f}_E > \alpha_n}} = \limv{n}\mu\bracsn{\norm{f}_E > \alpha}
+    \]
+
+    by \hyperref[continuity from below]{proposition:measure-properties}. 
+
+    (2): For each $\alpha > 0$ and $t \in (0, 1)$, 
+    \[
+    \bracsn{\norm{f + g}_E > \alpha} \subset \bracsn{\norm{f}_E > t\alpha} \cup \bracsn{\norm{g}_E > (1-t)\alpha}
+    \]
+
+    (3): For each $\alpha > 0$, $\bracsn{\norm{f}_E > \alpha} = 
+    
+    
+    Assume without loss of generality that $\fF \subset 2^{L^0(X; [0, \infty))}$ and $f \in L^0(X; [0, \infty))$. Fix $\alpha > 0$, then for each $t \in (0, 1)$, 
+    \[
+    \lambda_{f}(\alpha) \le \lambda_{f - g}(t\alpha) + \lambda_{g}((1 - t)\alpha)
+    \]
+\end{proof}
+
+
+