Lebesgue vs Riemann on the rationals

  • I
  • Thread starter BWV
  • Start date
  • #1
BWV
838
851
So Riemann integrals on ℚ can be <> 0 but Lebesgue integrals on ℚ all have measure zero?
 

Answers and Replies

  • #2
Office_Shredder
Staff Emeritus
Science Advisor
Gold Member
4,436
519
What is the definition of a Riemann integral on ##\mathbb{Q}##?
 
  • #3
BWV
838
851
What is the definition of a Riemann integral on ##\mathbb{Q}##?

so does the limit of Riemann sums not work inℚ? I was thinking it did, but not sure which is why I asked the question
 
  • #4
wrobel
Science Advisor
Insights Author
794
518
Lebesgue integrals on ℚ all have measure zero?
Integral is a number. It can not "have measure zero".
If say for a measure in ##\mathbb{Q}## you employ ##\delta(0)## then Lebesgue ##\int_\mathbb{Q} f=f(0)## for any ##f:\mathbb{Q}\to\mathbb{R}##
 
  • #5
mathman
Science Advisor
7,934
487
so does the limit of Riemann sums not work inℚ? I was thinking it did, but not sure which is why I asked the question
Riemann sums are usually defined as sums where an interval is partitioned into small intervals. You need to set up a definition. Also what is your definition of Lebesgue integral here?
 
  • #6
BWV
838
851
Riemann sums are usually defined as sums where an interval is partitioned into small intervals. You need to set up a definition. Also what is your definition of Lebesgue integral here?
Reading about the Lebesgue integral in the context of statistics and the examples of integration of ℚ = 0 like the common example of probability of picking a rational number from [0,1] in ℝ, but it seems an odd notion (i.e. I have something wrong) that integration only works in ℝ, or that Riemann works where Lebesgue does not
 
  • #7
331
129
Reading about the Lebesgue integral in the context of statistics and the examples of integration of ℚ = 0 like the common example of probability of picking a rational number from [0,1] in ℝ, but it seems an odd notion (i.e. I have something wrong) that integration only works in ℝ, or that Riemann works where Lebesgue does not
What source are you reading about the Lebesgue integral with statistics/probability? I suspect that you may be overgeneralizing from the examples of Lebesgue integration you've studied.

Say I looked at a few pages of a calculus book and saw that the three examples there were all definite integrals on the range [0,2]. Would it be correct for me to then assume that integrals are only ever done from 0 to 2?
 
  • #8
BWV
838
851
What source are you reading about the Lebesgue integral with statistics/probability? I suspect that you may be overgeneralizing from the examples of Lebesgue integration you've studied.

Say I looked at a few pages of a calculus book and saw that the three examples there were all definite integrals on the range [0,2]. Would it be correct for me to then assume that integrals are only ever done from 0 to 2?

its the first chapter of a Stochastic Calc for Finance book which is an intro to measure-theoretic probability. my takeaway was that Lebesgue measure -> Lebesgue integral, but confused as any Lebesgue measure on ℚ or ℤ is zero. The other option is Lebesgue integration requires a mapping from ℤ or ℚ to ℝ with measurable functions, like a step function for ℤ
 
  • #9
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
16,927
8,761
its the first chapter of a Stochastic Calc for Finance book which is an intro to measure-theoretic probability. my takeaway was that Lebesgue measure -> Lebesgue integral, but confused as any Lebesgue measure on ℚ or ℤ is zero. The other option is Lebesgue integration requires a mapping from ℤ or ℚ to ℝ with measurable functions, like a step function for ℤ
You're missing the point that the Riemann integral is defined only on the Real numbers. There is simply no definition of a Riemann integral on the rationals.

You may be confused with the Riemann integral of functions that are zero on the irrational numbers and non-zero on the rationals. The Riemann integral for these functions generally does not converge.
 
  • #10
BWV
838
851
Thanks, so calculus is only on the reals, but not sure technically why the Riemann integral could not work in ℚ for certain functions. For example, the Wikipedia article on Thomae's function:
1617201503594.png


claims that it is Riemann integrable
  • f
    is Riemann integrable on any interval and the integral evaluates to {\displaystyle 0}
    {\displaystyle 0}
    over any set.
The Lebesgue criterion for integrability states that a bounded function is Riemann integrable if and only if the set of all discontinuities has measure zero.[4] Every countable subset of the real numbers - such as the rational numbers - has measure zero, so the above discussion shows that Thomae's function is Riemann integrable on any interval. The function's integral is equal to 0 over any set because the function is equal to zero almost everywhere.


integrating something trivial like ##y=2## on ℚ would not have any discontinuities in ℚ (whereas ##y=1/2 \sqrt x ## would be discontinuous in ℚ)
 
  • #11
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
16,927
8,761
The function$$f(x)=\begin{cases} 1 & x \in \mathbb{N} \\ 0 & x \notin \mathbb{N} \end{cases}$$ is Riemann integrable. But, that's not the same as integration on the integers.
 
  • #12
Stephen Tashi
Science Advisor
7,642
1,496
Thanks, so calculus is only on the reals, but not sure technically why the Riemann integral could not work in ℚ for certain functions.

What do you mean by "not work"?
 
  • #14
Stephen Tashi
Science Advisor
7,642
1,496
The limit of Riemann sums would not exist

Then what is the relevance of the example of Thomae's function to your question?

We need to pay attention to the exact wording of definitions.

Most authors don't define "Riemann integral" for ##\int_S f(x) dx ## where ##S## is an arbitrary set. The Riemann integral is defined only when ##S## is a set that is an interval. The set ##S## must be a bounded interval unless we are also dealing with the definition of "improper" Riemann integrals.

So it isn't clear what one would mean by Riemann integrating ##f## "over" ## \mathbb{Q}##. By that phrase one might mean integration a function that is only defined on the set ##\mathbb{Q}## and doing the integration over the unbounded interval ##(-\infty, \infty)## in ##\mathbb{R}##.

There are examples of functions where the Riemann integral ##\int_{-\infty}^{\infty} f(x) dx = 0 ## and there exist values of ##f(x)## where ##f## is not zero. But such functions are not examples of where Riemann sums do not exist.
 
  • Informative
Likes PeroK and BWV
  • #15
Office_Shredder
Staff Emeritus
Science Advisor
Gold Member
4,436
519
I think a thing that's true is you could define a Riemann sum to be restricted to only sample rational numbers as the endpoints of each little interval, and then all the theory would work without any issues and a function like ##f(x)= 1## if ##x\in \mathbb{Q}## and 0 otherwise (or not even defined otherwise, whatever) would just have a well defined non zero integral.

Then I guess the question is can you pick a measure on ##\mathbb{Q}## that would give you the same result. It seems like a well founded and interesting question to me. My guess is if ##m## is the normal measure on ##\mathbb{R}##, that ##\tilde{m}(A) = m(\bar{A})## works, i.e. you take the closure of A in ##\mathbb{R}## and use that set's measure.
 

Related Threads on Lebesgue vs Riemann on the rationals

Replies
6
Views
6K
Replies
6
Views
2K
Replies
30
Views
8K
Replies
4
Views
7K
Replies
7
Views
18K
Replies
4
Views
3K
Replies
1
Views
2K
  • Last Post
Replies
9
Views
3K
  • Last Post
Replies
6
Views
7K
Replies
1
Views
2K
Top