Riemann integrability of functions with countably infinitely many dis-

  • #1
Eclair_de_XII
1,066
90
TL;DR Summary
-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.
 

Answers and Replies

  • #2
mathman
Science Advisor
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Do you have a question?
 
  • #3
Office_Shredder
Staff Emeritus
Science Advisor
Gold Member
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##|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?
 
  • #4
Eclair_de_XII
1,066
90
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.
 

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