Sandwich theorem for lebesgue integral

  • Context: Graduate 
  • Thread starter Thread starter onthetopo
  • Start date Start date
  • Tags Tags
    Integral Theorem
Click For Summary

Discussion Overview

The discussion revolves around the existence of a sandwich theorem for the Lebesgue integral, specifically addressing conditions under which a sequence of functions bounded by integrable functions also results in an integrable function. Participants explore the implications of convergence of sequences of functions and the conditions necessary for applying the Dominated Convergence Theorem (DCT).

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants inquire whether a sandwich theorem exists for Lebesgue integrals, questioning if the integrability of bounding functions implies the integrability of the function in between.
  • There is a discussion about the measurability of the functions involved, with some suggesting that if the middle function is not measurable, it complicates the argument.
  • One participant asserts that if the functions are nonnegative, the comparison theorem can be applied to show integrability.
  • Another participant mentions that if the bounding functions converge almost everywhere, then the integrals of the bounding functions converge to the integral of the limit function.
  • There is a challenge regarding the necessity of pointwise convergence of the middle function to a limit function for applying the DCT.
  • Some participants discuss the conditions under which the DCT can be applied, including the need for the sequences to converge and the implications of absolute convergence.
  • References to specific exercises and theorems in measure theory literature are made, suggesting that these results may be found in standard texts.

Areas of Agreement / Disagreement

Participants express differing views on the necessity of pointwise convergence and the conditions required for the DCT. There is no consensus on whether the sandwich theorem can be established under the discussed conditions, and the discussion remains unresolved regarding the implications of the various assumptions.

Contextual Notes

Participants highlight the importance of measurability and the nature of convergence in the context of the Lebesgue integral, indicating that assumptions about the functions' properties are crucial to the discussion.

onthetopo
Messages
35
Reaction score
0
It would be helpful if there is a sandwich theorem for lebesgue integreal. Does it exist?
Ie. if fn<=hn<=gn on measure space (X,S,u) and fn and hn are integreable and measurable , then
g must also be in L1.

A related claim is that if fn-> f and hn->h, does gn also necessarily converge to some g?
 
Last edited:
Physics news on Phys.org
Did you mean to ask if h_n (not g) is in L^1? What do you know about h_n - in particular, is it measurable? If it's just an arbitrary function then it's easy to come up with examples of nonmeasurable functions sandwiched between L^1 functions...

On the other hand, if h_n is measurable, then it is L^1 by the monotonicity of the integral (you have to be a little careful here).

And when you write f_n -> f, what notion of convergence are you using?

For future reference, you should always post your ideas when you ask for help. Forum rules and such.
 
morphism said:
Did you mean to ask if h_n (not g) is in L^1? What do you know about h_n - in particular, is it measurable? If it's just an arbitrary function then it's easy to come up with examples of nonmeasurable functions sandwiched between L^1 functions...

On the other hand, if h_n is measurable, then it is L^1 by the monotonicity of the integral (you have to be a little careful here).

And when you write f_n -> f, what notion of convergence are you using?

For future reference, you should always post your ideas when you ask for help. Forum rules and such.

Sorry , you are right.
Ideally, if fn, gn, and hn are all nonnegative, we can use comparison theorem to show that hn is in L1. But how about the case that fn, gn, hn are simply any measurable functions?
 
Any measurable function... is it integrable? If we have fn<=gn<=hn with fn and hn integrable, then |fn| and |hn| are integrable as well, and hence max(|fn|,|hn|) is also. What can you say about |gn| compared to this?
 
The question should include fn, hn, f,h integreable
then max(|fn|,|hn|) is also integreable,
then |gn|<x(|fn|,|hn|), thus gn is in L1,
But still it doesn't say "integral gn" converges to "integral g" for some g
 
That last bit follows from the dominated convergence theorem.
 
To use DCT, we need to show there exists a g such that g->gn
but this is true, since |gn|<|fn|+|hn|<|f|+|h|+2epsilon
thus |gn| converges to |f|+|h|
Now, suppose the space is complete, then every absolutely convergent sequence is convergent, thus there exists a g such that gn->g
is this right?
 
Hold on - no. If you don't know that g_n converges in the first place, then you can't say anything.

On the other hand, we do have the following result: Let {f_n} and {h_n} be sequences of integrable functions that converge a.e. to f and h (which are also integrable functions). Let {g_n} be a sequence of measurable functions such that f_n <= g_n <= h_n for all n, and such that g_n -> g pointwise. If [itex]\int f_n \to \int f[/itex] and [itex]\int h_n \to \int h[/itex], then [itex]\int g_n \to \int g[/itex].

I don't know off the top of my head if any of the conditions can be relaxed on this. (In fact, they might even require some strengthening!)
 
morphism said:
Hold on - no. If you don't know that g_n converges in the first place, then you can't say anything.

On the other hand, we do have the following result: Let {f_n} and {h_n} be sequences of integrable functions that converge a.e. to f and h (which are also integrable functions). Let {g_n} be a sequence of measurable functions such that f_n <= g_n <= h_n for all n, and such that g_n -> g pointwise. If [itex]\int f_n \to \int f[/itex] and [itex]\int h_n \to \int h[/itex], then [itex]\int g_n \to \int g[/itex].

I don't know off the top of my head if any of the conditions can be relaxed on this. (In fact, they might even require some strengthening!)

Wonderful, How did you prove this? if too long, what reference book did you find this theorem?
 
  • #10
I did an exercise of this sort back when I took measure theory (so I might have gotten some of the details wrong). If I'm not mistaken, this might also be an exercise in Royden (in the chapter that contains the Lebesgue dominated convergence theorem for the real line; chapter 3?). Anyway, the idea is to use the dominated convergence theorem. It might also be helpful to consider the case f_n=-g_n (so |h_n| <= f_n) first.
 
  • #11
why do we need gn->g pointwise
|gn|<|fn|+|hn|<|f|+|h|+2epsilon
thus |gn| converges to |f|+|h|
 
Last edited:

Similar threads

  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 33 ·
2
Replies
33
Views
7K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 4 ·
Replies
4
Views
7K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K