Update Math4121_L35.md

pages/Math4121/Math4121_L35.md

@@ -89,5 +89,26 @@ $$
 
 Proof:
 
+First to show the limit exists almost everywhere. It suffices to show
+
+$$
+\mathcal{U}=\{x\in E: f_n(x) \text{ is unbounded}\}
+$$
+
+has measure 0.
+
+Let $\epsilon>0$ and write
+
+$$
+U=\bigcup_{n=1}^{\infty} E_n
+$$
+
+where $E_n=\{x\in E: |f_n(x)|\geq \epsilon\}$.
+
+Then $U\subseteq \mathcal{U}$ and $m(U)<\epsilon$.
+
+
+CONTINUE NEXT TIME.
+
 QED