Riemann integrability of functions with countably infinitely many dis-

[Eclair_de_XII]

TL;DR

-continuities.

Define $f:[a,b]\longrightarrow \mathbb{R}$. Assume that $f$ is non-negative 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 $\Delta t=t_2-t_1$ also. Then the following must hold:

\begin{eqnarray}
M\Delta t - m\Delta t &=&(M-m)\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.

 

[mathman]

Do you have a question?

 

[Office_Shredder]

$|f(x)-f(a_m)|$ can certainly be larger than the supremum of $f$, for example on $[-1,1]$ the function $f(x)=x$ has supremum $1$, but $|f(-1)-f(1)| =2$. That said your claim that the function is bounded is still true, you just need to think a little more about why.

I also don't see where in the rest of your proof you use the fact that the $g_m$s are bounded, is there a piece missing?

 

[Eclair_de_XII]

Do you have a question?

I'm sorry for not clarifying in the original post. I was asking for feedback on my proof. And if possible, I want to ask if the fourth paragraph where I define an alias for the supremum of $G$ sounds too informal.

the function has supremum , but .

This proof is for functions that takes on non-negative values. But if I choose to extend the family of functions to which this theorem might apply, I would replace $\sup f$ with $|\sup f - \inf f|$.

is there a piece missing?

It's essential to proving that $\alpha$ as declared in the fourth paragraph exists.