Update Math4121_L12.md

pages/Math4121/Math4121_L12.md

@@ -96,4 +96,44 @@ Define $\epsilon_n=\sup_{x\in[a,b]}|f_n(x)-f(x)|$.
 
 By uniform convergence, $\epsilon_n\to 0$ as $n\to\infty$.
 
-CONTINUE HERE
+$$
+f_n(x)-\epsilon_n\leq f(x)\leq f_n(x)+\epsilon_n
+$$
+
+$$
+\int_a^b f_n-\epsilon_nd\alpha\leq\overline{\int_a^b}fd\alpha\leq\int_a^bf_n+\epsilon_nd\alpha
+$$
+
+
+$$
+\int_a^b f_n-\epsilon_n d\alpha\leq\underline{\int_a^b}fd\alpha\leq\int_a^b f_n+\epsilon_n d\alpha
+$$
+
+So,
+
+$$
+0\leq\overline{\int_a^b}fd\alpha-\underline{\int_a^b}fd\alpha\leq\int_a^b \epsilon_n d\alpha+\int_a^b \epsilon_n d\alpha=2\epsilon_n[\alpha(b)-\alpha(a)]
+$$
+
+So $f\in\mathscr{R}(\alpha)$ and
+
+$$
+\int_a^b fd\alpha\leq \int_a^b f_n d\alpha+\int_a^b \epsilon_n d\alpha\leq \int_a^b fd\alpha+2\epsilon_n d\alpha
+$$
+
+So,
+
+$$
+\int_a^b fd\alpha-\int_a^b \epsilon_n d\alpha\leq \int_a^b f_n d\alpha\leq \int_a^b fd\alpha+\int_a^b \epsilon_n d\alpha
+$$
+
+Since $\int_a^b \epsilon_n d\alpha\to 0$ as $n\to\infty$, by the squeeze theorem, we have
+
+by the squeeze theorem, we have
+
+$$
+\lim_{n\to\infty}\int_a^b f_nd\alpha=\int_a^b fd\alpha
+$$
+
+_Key is that $\int_a^b (f-f_n)d\alpha\leq \sup_{x\in[a,b]}|f-f_n|(\alpha(b)-\alpha(a))$_
+