diff --git a/src/fa/lp/ui.tex b/src/fa/lp/ui.tex
index b00c919..fc15291 100644
--- a/src/fa/lp/ui.tex
+++ b/src/fa/lp/ui.tex
@@ -156,3 +156,105 @@
 \begin{proof}
     By \autoref{theorem:vitali-convergence}. 
 \end{proof}
+
+\begin{lemma}[Scheffé]
+\label{lemma:scheffe}
+    Let $(X, \cm, \mu)$ be a measure space, $p \in [1, \infty)$, $E$ be a normed vector space over $K \in \RC$, $\fF \subset 2^{L^p(X; E)}$ be a filter, and $g \in L^p(X; E)$, then $\fF \to f$ in $L^p(X; E)$ if and only if:
+    \begin{enumerate}
+        \item[(M)] $\fF \to g$ is locally in measure. 
+        \item[(N)] $\lim_{f, \fF} \norm{f}_{L^p(X; E)} = \norm{g}_{L^p(X; E)}$. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    It is sufficient to show conditions (UI) and (T) of the \hyperref[Vitali Convergence Theorem]{theorem:vitali-convergence}. 
+
+    (T): Let $\eps > 0$. By (N), there exists $F_1 \in \fF$ such that $|\norm{f}_{L^p(X; E)}^p - \norm{g}_{L^p(X; E)}^p| < \eps$ for all $f \in F_1$. By the \hyperref[Dominated Convergence Theorem]{theorem:dct}, there exists $A \in \cm$ with $\mu(A) < \infty$ such that:
+    \begin{enumerate}[label=(\roman*)]
+        \item $\int_{A^c} \norm{g}_E^p < \eps$. 
+    \end{enumerate}
+    
+    Since $\norm{g}_E^p d\mu \ll \mu$, by \autoref{proposition:ac-epsilon-delta}, there exists $\delta > 0$ such that:
+    \begin{enumerate}[label=(\roman*), start=1]
+        \item For each $B \in \cm$ with $\mu(B) < \delta$, $\int_{B}\norm{g}_E^p d\mu < \eps$. 
+        \item $\int_{A}[(\norm{g}_E - \delta) \vee 0]^p d\mu > \int_A \norm{g}_E^p d\mu - \eps$. 
+    \end{enumerate}
+
+    By (M), there exists $F_2 \in \fF$ with $F_2 \subset F_1$ such that for each $f \in F_2$, $\mu(A \cap \bracs{\norm{f - g}_E > \delta}) < \delta$. In which case, for any $f \in F_2$, 
+    \begin{align*}
+        \int_A \norm{f}_E^p d\mu &\ge \int_{A \cap \bracs{\norm{f - g}_E \le \delta}}\norm{f}_E^p d\mu \\
+        &\ge \int_{A}[(\norm{g}_E - \delta) \vee 0]^p d\mu - \int_{A \cap \bracs{\norm{f - g}_E > \delta}}\norm{g}_E^p d\mu
+    \end{align*}
+
+    By (iii), 
+    \[
+    \int_A \norm{f}_E^p d\mu \ge \int_A \norm{g}_E^p d\mu - \eps - \int_{A \cap \bracs{\norm{f - g}_E > \delta}}\norm{g}_E^p d\mu
+    \]
+
+    and by (ii), $\int_A \norm{f}_E^p d\mu \ge \int_A\norm{g}_E^p d\mu - 2\eps$. Finally, by (i), 
+    \[
+    \int_A \norm{f}_E^p d\mu \ge \int \norm{g}_E^p d\mu - 3\eps \ge \int \norm{f}_E^p d\mu - 4\eps
+    \]
+
+    and $\int_{A^c}\norm{f}_E^p d\mu \le 4\eps$. 
+    
+    
+    (UI): Let $\eps > 0$. By (N) and (T), there exists $F_1 \in \fF$ and $A \in \cm$ with $\mu(A) < \infty$ such that for every $f \in F_1$, 
+    \begin{enumerate}[label=(\roman*)]
+        \item $|\norm{f}_{L^p(X; E)}^p - \norm{g}_{L^p(X; E)}^p| < \eps$. 
+        \item $\int_{A^c}\norm{f}_E^p d\mu < \eps$ and $\int_{A^c}\norm{g}_E^p d\mu$. 
+    \end{enumerate}
+    
+    By (i) and (ii), 
+    \begin{enumerate}[label=(\roman*), start=2]
+        \item $\abs{\int_{A}\norm{f}_E^p d\mu - \int_{A}\norm{g}_E^p} \le 3\eps$. 
+    \end{enumerate}
+    
+
+    Since $\norm{g}_E^p d\mu \ll \mu$, by \autoref{proposition:ac-epsilon-delta}, there exists $\delta > 0$ such that:
+    \begin{enumerate}[label=(\roman*), start=3]
+        \item For each $B \in \cm$ with $\mu(B) < \delta$, $\int_{B}\norm{g}_E^p d\mu < \eps$. 
+        \item $\int_{A}[(\norm{g}_E - \delta) \vee 0]^p d\mu > \int_A \norm{g}_E^p d\mu - \eps$. 
+    \end{enumerate}
+
+    By (M), there exists $F_2 \in \fF$ with $F_2 \subset F_1$ such that for every $f \in F_2$,  
+    \begin{enumerate}[label=(\roman*), start=5]
+        \item $\mu(A \cap \bracs{\norm{f - g}_E > \delta}) < \delta$. 
+    \end{enumerate}
+    
+    Let $B \in \cm$ with $B \subset A$ and $\mu(B) < \delta$, then for any $f \in F_3$, 
+    \begin{align*}
+        \int_{A \setminus B}\norm{f}_E^p d\mu &\ge \int_{(A \setminus B) \cap \bracs{\norm{f - g}_E \le \delta}}\norm{f}_E^p d\mu \\
+        &\ge \int_{(A \setminus B) \cap \bracs{\norm{f - g}_E \le \delta}}(\norm{g}_E - \delta \vee 0)^pd\mu \\
+        &\ge \int_{(A \setminus B)}(\norm{g}_E - \delta \vee 0)^pd\mu - \int_{\bracs{\norm{f - g}_E > \delta}} \norm{g}_E^p d\mu
+    \end{align*}
+
+    By (vi) and (iv), $\int_{\bracs{\norm{f - g}_E > \delta}} \norm{g}_E^p d\mu < \eps$, so
+    \[
+    \int_{A \setminus B}\norm{f}_E^p d\mu \ge \int_{(A \setminus B)}(\norm{g}_E - \delta \vee 0)^pd\mu - \eps
+    \]
+
+    By (v), 
+    \[
+    \int_{A \setminus B}\norm{f}_E^p d\mu \ge \int_{(A \setminus B)}\norm{g}_E^pd\mu - 2\eps
+    \]
+
+    Since $\mu(B) < \delta$, by (iv),
+    \[
+    \int_{A \setminus B}\norm{f}_E^p d\mu \ge \int_{A}\norm{g}_E^pd\mu - 3\eps
+    \]
+
+    and by (iii), 
+    \[
+    \int_{A \setminus B}\norm{f}_E^p d\mu \ge \int_{A}\norm{f}_E^pd\mu - 6\eps
+    \]
+
+    so $\int_{B}\norm{f}_E^p d\mu \le 6\eps$ for all $B \in \cm$ with $B \subset A$ and $\mu(B) < \delta$.
+    
+    Finally, by (i) and \hyperref[Markov's Inequality]{theorem:markov-inequality}, there exists $M \ge 0$ such that $\mu\bracs{\norm{f}_E \ge M} < \delta$ for all $f \in F_3$. Therefore for any $f \in F_3$, 
+    \[
+        \int_{\bracs{\norm{f}_E \ge M}}\norm{f}_E^p d\mu \le \int_{A \cap \bracs{\norm{f}_E \ge M}}\norm{f}_E^p d\mu + \int_{A^c}\norm{f}_E^p d\mu \le 7\eps
+    \]
+    
+    
+\end{proof}
+