diff --git a/src/measure/measure/lebesgue-stieltjes.tex b/src/measure/measure/lebesgue-stieltjes.tex
index 3879a25..8501e5f 100644
--- a/src/measure/measure/lebesgue-stieltjes.tex
+++ b/src/measure/measure/lebesgue-stieltjes.tex
@@ -46,7 +46,17 @@
         \item[(U)] For any function $G: \real \to \real$ satisfying (1), $F - G$ is constant.
     \end{enumerate}
 
-    Conversely, if $F: \real \to \real$ is a Stieltjes function, then there exists a unique Borel measure $\mu_F: \cb_\real \to [0, \infty]$ satisfying (1), and $\mu_F$ is the \textbf{Lebesgue-Stieltjes measure} associated with $F$.
+    Conversely, if $F: \real \to \real$ is a Stieltjes function, then 
+    \begin{enumerate}
+        \item[(1C)] There exists a unique Borel measure $\mu_F: \cb_\real \to [0, \infty]$ such that for all $-\infty < a < b < \infty$, 
+        \[
+        \mu_F((a, b]) = F(b) - F(a)
+        \]
+    \end{enumerate}
+    
+    In which case, $\mu_F$ is the \textbf{Lebesgue-Stieltjes measure} associated with $F$.
+    
+    
 \end{definition}
 \begin{proof}[Proof {{\cite[Theorem 1.16]{Folland}}}. ]
     (1): Let
@@ -71,7 +81,7 @@
     (U): If $G: \real \to \real$ satisfies (1), then $F(x) = G(x) - G(0)$. Hence $F - G$ is constant.
 
 
-    Converse: Let $\ci = \bracs{(a, b]| -\infty < a \le b < \infty}$ be the elementary family of h-intervals and
+    (1C): Let $\ci = \bracs{(a, b]| -\infty < a \le b < \infty}$ be the elementary family of h-intervals and
     \[
     \alg = \bracs{\bigsqcup_{j = 1}^n (a_j, b_j] \bigg | \seqf{(a_j, b_j]} \subset \ci, n \in \natp}
     \]