Pointwise convergence of Riemann integrable functions

In summary, changing the order of limit and integration in Riemann integration is possible if the functions are not only pointwise convergent but also bounded. This can be done using the dominated convergence theorem for Lebesgue measurable functions, which is an analog of the theorem for Riemann integrable functions. However, it is important to note that the limit function of a sequence of Riemann integrable functions may not be Riemann integrable itself, leading to the non-existence of the Riemann integral of the limit function. This can be seen in the example of a bounded sequence of Riemann integrable functions on a closed interval that converges pointwise to a bounded but not Riemann integrable function.
  • #1
bobby2k
127
2
Hello

Normally in order to change the order of limit and integration in rimann integration, you need uniform convergence.

But let's say that you are not able to prove uniform convergence, but only pointwise convergence. And let's say you are able to prove that the functions are also bounded. Could you then use something like the dominated convergence theorem for lebesgue measurable functions? The problem is offcourse that it holds for integrals defined another way.

But let's say you have a sequence of riemann integrable functions, that converge pointwise to a function, and all the functions are bounded, can we then change the order of limit and integration?
 
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  • #2
bobby2k said:
Hello

Normally in order to change the order of limit and integration in rimann integration, you need uniform convergence.

But let's say that you are not able to prove uniform convergence, but only pointwise convergence. And let's say you are able to prove that the functions are also bounded. Could you then use something like the dominated convergence theorem for lebesgue measurable functions? The problem is offcourse that it holds for integrals defined another way.

But let's say you have a sequence of riemann integrable functions, that converge pointwise to a function, and all the functions are bounded, can we then change the order of limit and integration?

Most of the time. To use bounded convergece as dominant convergence the domain of integration has to be finite.

Simple counterexample for infinite domain.

f_n(x) = 1 for n<x<n+1, =0 otherwise.
functions converge to 0, but integral remains 1.
 
  • #3
mathman said:
Most of the time. To use bounded convergece as dominant convergence the domain of integration has to be finite.

Simple counterexample for infinite domain.

f_n(x) = 1 for n<x<n+1, =0 otherwise.
functions converge to 0, but integral remains 1.

Yeah, but what if we assume that it is finite domain? I mean technically the riemann integral is only defined on a bounded and closed interval?, but you can take indefinite integrals, but let's assume you don't do that in this case.
 
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  • #4
bobby2k said:
I mean technically the riemann integral is only defined on a bounded and closed interval?

Not true. Riemann sums can be defined over unbounded closed intervals of ##\overline{\mathbb{R}}##.
 
  • #5
bobby2k said:
But let's say that you are not able to prove uniform convergence, but only pointwise convergence. And let's say you are able to prove that the functions are also bounded. Could you then use something like the dominated convergence theorem for lebesgue measurable functions?
Yes there is an analog of dominated convergence for Riemann integrable functions but it is decidedly less useful. Suppose ## f_n: [a,b] \to \mathbb{R} ## and for some M we have that ##|f_n(x)| < M ## for each n. If ##f_n \to f ## pointwise and ##f## is riemann integrable, then you may interchange limits and integrals.
 
  • #6
Jorriss said:
Yes there is an analog of dominated convergence for Riemann integrable functions but it is decidedly less useful. Suppose ## f_n: [a,b] \to \mathbb{R} ## and for some M we have that ##|f_n(x)| < M ## for each n. If ##f_n \to f ## pointwise and ##f## is riemann integrable, then you may interchange limits and integrals.
This is a simple consequence of the corresponding theorem for Lebesgue integrals, since a Riemann integrable function is also Lebesgue integrable, and its Riemann integral is the same as its Lebesgue integral.

It may happen, however, that the limit function of a sequence of Riemann integrable functions on a closed interval exists and is bounded, but not Riemann integrable, so that the Riemann integral of the limit function does not exist.

As an example, let q1, q2, q3,... be an enumeration of the rational numbers in [0,1].
For each n, let fn:[0,1]→[0,1] be defined fn(x)=1 if x=qi for some i≤n, and fn(x)=0 otherwise. Each fn is Riemann integrable on [0,1] with Riemann integral 0.
But fn → f, pointwise, where f:[0,1]→[0,1] is defined as f(x)=1 if x is rational and f(x)=0 otherwise. f is bounded but not Riemann integrable on [0,1].
 
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  • #7
Jorriss said:
Yes there is an analog of dominated convergence for Riemann integrable functions but it is decidedly less useful. Suppose ## f_n: [a,b] \to \mathbb{R} ## and for some M we have that ##|f_n(x)| < M ## for each n. If ##f_n \to f ## pointwise and ##f## is riemann integrable, then you may interchange limits and integrals.



Erland said:
This is a simple consequence of the corresponding theorem for Lebesgue integrals, since a Riemann integrable function is also Lebesgue integrable, and its Riemann integral is the same as its Lebesgue integral.

It may happen, however, that the limit function of a sequence of Riemann integrable functions on a closed interval exists and is bounded, but not Riemann integrable, so that the Riemann integral of the limit function does not exist.

As an example, let q1, q2, q3,... be an enumeration of the rational numbers in [0,1].
For each n, let fn:[0,1]→[0,1] be defined fn(x)=1 if x=qi for some i≤n, and fn(x)=0 otherwise. Each fn is Riemann integrable on [0,1] with Riemann integral 0.
But fn → f, pointwise, where f:[0,1]→[0,1] is defined as f(x)=1 if x is rational and f(x)=0 otherwise. f is bounded but not Riemann integrable on [0,1].

Thanks!
 
  • #8
pwsnafu said:
Not true. Riemann sums can be defined over unbounded closed intervals of ##\overline{\mathbb{R}}##.

You are certainly free to define Riemann sums/integrals in whatever way pleases you, but the standard definitions of those terms do not apply to unbounded intervals in any meaningful way.

You can definitely talk about improper Riemann integrals, which are typically defined in terms of limits of Riemann integrals and not in terms of Riemann sums, or Lebesgue integrals, which look kinda like Riemann integrals (but not really) when applied to compact intervals, both of which are perfectly well-defined on all intervals. And there are other way to define integrals that are "Riemann-like" and apply to functions defined on unbounded intervals. But the standard definitions define Riemann integrals as limits of Riemann sums, which are defined using finite partitions of compact intervals.
 

1. What is pointwise convergence of Riemann integrable functions?

Pointwise convergence of Riemann integrable functions is a type of convergence in which a sequence of functions converges to a limit function at each point in the domain. This means that for any given point in the domain, the sequence of function values will eventually become arbitrarily close to the value of the limit function.

2. What is the difference between pointwise convergence and uniform convergence?

The main difference between pointwise convergence and uniform convergence is that in pointwise convergence, the convergence occurs at each point in the domain separately, while in uniform convergence, the convergence occurs uniformly across the entire domain. This means that in uniform convergence, the sequence of functions approaches the limit function at the same rate at every point in the domain.

3. How is pointwise convergence related to Riemann integrability?

Pointwise convergence of Riemann integrable functions implies that the limit function is also Riemann integrable. This means that the limit of a sequence of Riemann integrable functions is also Riemann integrable, and the integral of the limit function is equal to the limit of the integrals of the sequence of functions.

4. Can a sequence of Riemann integrable functions converge pointwise but not uniformly?

Yes, it is possible for a sequence of Riemann integrable functions to converge pointwise but not uniformly. This can occur when the functions in the sequence have different rates of convergence at different points in the domain, causing the convergence to not be uniform across the entire domain.

5. How is pointwise convergence related to continuity?

If a sequence of continuous functions converges pointwise to a limit function, then the limit function is also continuous. However, the converse is not always true - a sequence of discontinuous functions can still converge pointwise to a continuous limit function. This is because pointwise convergence only considers the behavior of the functions at each individual point, rather than the behavior across the entire domain.

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