Riemann integrability of functions with countably infinitely many dis-

In summary: Also, I use the fact that ##g_m## is bounded in the sixth paragraph where I set ##M=\sup f(B(y_0,\delta))## and ##m=\inf f(B(y_0,\delta))## and use the fact that ##g_m## is bounded to show that ##M-m\leq\frac{\epsilon}{2}##. This then allows me to choose a partition that satisfies the desired condition.In summary, the conversation discusses a proof that shows the convergence of upper and lower sums of a function with respect to a specific partition. The proof involves defining a sequence of bounded functions and using the fact that the supremum of the set of these functions exists. The conversation also mentions extending the theorem
  • #1
Eclair_de_XII
1,083
91
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.
 
Physics news on Phys.org
  • #2
Do you have a question?
 
  • #3
##|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
mathman said:
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.

Office_Shredder said:
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|##.

Office_Shredder said:
is there a piece missing?
It's essential to proving that ##\alpha## as declared in the fourth paragraph exists.
 

1. What is Riemann integrability?

Riemann integrability is a mathematical concept that refers to the ability to find the definite integral of a function over a given interval. It is based on the idea of partitioning the interval into smaller subintervals and calculating the area under the curve using rectangles.

2. What does it mean for a function to have countably infinitely many discontinuities?

A function with countably infinitely many discontinuities means that there are infinitely many points in the domain of the function where the function is not continuous. This means that there are abrupt changes or jumps in the function's values at these points.

3. How does the presence of countably infinitely many discontinuities affect the Riemann integrability of a function?

If a function has countably infinitely many discontinuities, it may not be Riemann integrable. This is because the Riemann integral is defined as the limit of a sum of rectangles, and the presence of discontinuities can cause the sum to become infinite or undefined.

4. Can a function with countably infinitely many discontinuities still be Lebesgue integrable?

Yes, a function with countably infinitely many discontinuities can still be Lebesgue integrable. Unlike the Riemann integral, the Lebesgue integral does not depend on the continuity of the function, but rather on its measurability. This means that as long as the function is measurable, it can be Lebesgue integrable.

5. Are there any techniques or methods for determining the Riemann integrability of a function with countably infinitely many discontinuities?

There are some techniques and methods that can be used to determine the Riemann integrability of a function with countably infinitely many discontinuities. These include the use of Lebesgue's criterion, which states that a function is Riemann integrable if and only if its set of discontinuities has measure zero. Additionally, the use of special types of partitions, such as the Darboux partition, can also help in determining the Riemann integrability of a function with discontinuities.

Similar threads

Replies
7
Views
1K
Replies
2
Views
1K
Replies
16
Views
2K
Replies
3
Views
2K
Replies
5
Views
381
Replies
1
Views
931
Replies
11
Views
980
Replies
1
Views
1K
Replies
14
Views
1K
Replies
24
Views
2K
Back
Top