 #1
Eclair_de_XII
 1,066
 90
 TL;DR Summary

continuities.
Define ##f:[a,b]\longrightarrow \mathbb{R}##. Assume that ##f## is nonnegative and bounded. Suppose there exists a point ##y_0\in [a,b]## at which ##f## fails to be continuous and suppose also that there exists a sequence of points ##a_n\in[a,b]## that converge to ##y_0##, where ##f## fails to be continuous at each ##a_n##. Then ##f## is integrable.
Assume any function ##f## with only finitely many discontinuities is integrable.
We show that there is a partition s.t. the upper sum and the lower sum of ##f## w.r.t. this partition converge onto one another.
Let ##\epsilon>0##.
Define a sequence of functions ##g_n:[a,b]\setminus(\{a_n\}_{n\in\mathbb{N}}\cup\{y_0\})## s.t. ##g_n(x)=f(x)f(a_n)##. Suppose there is a ##g_m## that is not bounded. Then there must exist a point ##x'## s.t. ##f(x')f(a_m)>\sup f[a,b]##. This is a contradiction. Hence, each ##g_n## is bounded, and moreover, ##G:=\{g_n(x):x\in\textrm{dom}(g_n)\}_{n\in\mathbb{N}}## is bounded.
Set ##\alpha=\sup G## and set ##\delta=\frac{1}{4}\cdot\frac{\epsilon}{\alpha}##. Now define points ##t_1,t_2## to be the endpoints of ##B(y_0,\delta)##. Set ##M=\sup f(B(y_0,\delta))## and ##m=\inf f(B(y_0,\delta))##; set ##\Delta t=t_2t_1## also. Then the following must hold:
\begin{eqnarray}
M\Delta t  m\Delta t &=&(Mm)\Delta t \\
&\leq&\alpha\Delta t \\
&=&2\delta\alpha\\
&=&2\left(\frac{1}{4}\cdot\frac{\epsilon}{\alpha}\right)\cdot\alpha \\
&=&\frac{\epsilon}{2}
\end{eqnarray}
Choose a partition ##P:=\{x_0,\ldots,x_n\}## for ##[a,b]\setminus B(y_0,\delta)## s.t. ##U(P,f)L(P,f)<\frac{\epsilon}{2}##. Now refine the partition with the points ##t_i## as described above in order to yield the result.
Let ##\epsilon>0##.
Define a sequence of functions ##g_n:[a,b]\setminus(\{a_n\}_{n\in\mathbb{N}}\cup\{y_0\})## s.t. ##g_n(x)=f(x)f(a_n)##. Suppose there is a ##g_m## that is not bounded. Then there must exist a point ##x'## s.t. ##f(x')f(a_m)>\sup f[a,b]##. This is a contradiction. Hence, each ##g_n## is bounded, and moreover, ##G:=\{g_n(x):x\in\textrm{dom}(g_n)\}_{n\in\mathbb{N}}## is bounded.
Set ##\alpha=\sup G## and set ##\delta=\frac{1}{4}\cdot\frac{\epsilon}{\alpha}##. Now define points ##t_1,t_2## to be the endpoints of ##B(y_0,\delta)##. Set ##M=\sup f(B(y_0,\delta))## and ##m=\inf f(B(y_0,\delta))##; set ##\Delta t=t_2t_1## also. Then the following must hold:
\begin{eqnarray}
M\Delta t  m\Delta t &=&(Mm)\Delta t \\
&\leq&\alpha\Delta t \\
&=&2\delta\alpha\\
&=&2\left(\frac{1}{4}\cdot\frac{\epsilon}{\alpha}\right)\cdot\alpha \\
&=&\frac{\epsilon}{2}
\end{eqnarray}
Choose a partition ##P:=\{x_0,\ldots,x_n\}## for ##[a,b]\setminus B(y_0,\delta)## s.t. ##U(P,f)L(P,f)<\frac{\epsilon}{2}##. Now refine the partition with the points ##t_i## as described above in order to yield the result.