Not riemann integrable => not lebesgue integrable ?

  • #1
"not riemann integrable" => "not lebesgue integrable"??

Hi,

In general if a function is not Riemann integrable does this mean the function is also not Lebesgue integrable? Why or why not?

I know that if the the function is Riemann integrable then its Lebesgue integrable, but I can't find any info about non-Riemann integrable functions. Maybe I'm not thinking about this in the right way...

For example, if you wanted to show 1/x is not Lebesgue integrable it would be very convenient if all you needed to show was that the Riemann integral was infinite. Otherwise I'm not sure how to compute the Lebesgue integral...Is there a straightforward method to computing integrals of such simple functions?

Any thoughts would be appreciated!
 

Answers and Replies

  • #2
Landau
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In general if a function is not Riemann integrable does this mean the function is also not Lebesgue integrable? Why or why not?
No. Because of the (standard) example: the characteristic function of the rationals. It is Lebesgue integrable (the Lebesgue measure of Q is zero), but not Riemann integrable (the lower integral is 0 and the upper integral is 1).
 
  • #3
AlephZero
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In general if a function is not Riemann integrable does this mean the function is also not Lebesgue integrable? Why or why not?
No. For example
f(x) = 1 when x = 1/N for any integer N
f(x) = 0 otherwise.

f(x) is not Riemann integrable but it is Lebesgue integrable.
 
  • #4


Right, ok. I guess I was thinking about this in a specific way.

Let's see. Are their conditions that can be set to change this? For example if a function is continuous almost everywhere on a domain but not Riemann integrable, can it be Lebesgue integrable?
 
  • #5
Landau
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For example if a function is continuous almost everywhere on a domain but not Riemann integrable, can it be Lebesgue integrable?
A bounded function on a compact interval is Riemann integrable if and only if it is continuous almost everywhere.
 
  • #6


I do recall that theorem. But what if the function is not bounded? For example, define f to be 1/x on (0,1] and 0 at 0. Then clearly f is not Riemann integrable. It would seem intuitively true that its not Lebesgue integrable.
 
  • #7
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No. For example
f(x) = 1 when x = 1/N for any integer N
f(x) = 0 otherwise.

f(x) is not Riemann integrable but it is Lebesgue integrable.
Are you sure that that is not Riemann integrable?

EDIT: to the OP, concerning your last post: indeed, and you can use it to prove that 1/x is not Lebesgue-integrable:
Define [tex]\chi_\epsilon[/tex] to be the characteristic function of [epsilon,1], i.e. the function that is 1 if x is in [epsilon,1], zero elsewhere.
It is obvious that, letting f(x) = 1/x, that the product [tex]f(x) \chi_\epsilon (x)[/tex] is Riemann-integrable and thus also Lebesgue-integrable. Now one of the most fundamental and basic theorems concerning Lebesgue-integration is the monotone convergence theorem: http://en.wikipedia.org/wiki/Monotone_convergence_theorem#Lebesgue_monotone_convergence_theorem
Lebesgue-integration is in a sense actually defined such that this theorem is true.
As you can verify, the sequence [tex]f \chi_{\epsilon_n}[/tex] with [tex]\epsilon_n := 1/n[/tex], fullfills the requirements of the theorem, such that [tex]\int_{Lebesgue} f(x) \mathrm dx = \int_{Lebesgue} \lim_n f(x) \chi_{\epsilon_n} \mathrm d x = \lim_n \int_{Lebesgue} f(x) \chi_{\epsilon_n} \mathrm d x = \lim_n \int_{Riemann} f(x) \chi_{\epsilon_n} \mathrm d x = \lim_n \ln{1} - \ln{\epsilon} = + \infty[/tex], and by definition we say that if the Lebesgue integral is infinite, the function is not integrable.

NOTE: detail: you might wonder why I was allowed to write down [tex]\int_{Lebesgue} f(x) \mathrm dx[/tex] (before starting all the equalities) because the integral might not have been defined (note: an integral equal to infinity is defined, in a sense that we can work with it, but for example, the Lebesgue-integral (over the whole real line) of [tex]\sin{x}/x[/tex] really is not defined). The thing is: f is a positive measurable function, and for such functions we can always write down the integral (it's a theorem), the only thing is, the integral might be infinite (as was the case here).

NOTE II: to list the idea used in each equality in that last long string of equalities:
first: pointwise limit
second: monotone convergence theorem
third: as stated in your first post: Riemann-integrable functions are Leb-int and their integrals coincide

EDIT II: this proof also answers another question of yours: one of the main methods of evaluating a lebesgue integral is writing it as a limiting case of Riemann-integrals (by use of limit-switching theorems)
 
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  • #8
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If I'm not mistaken, the integral

[tex]\int_0^{+\infty}{\frac{\sin(x)}{x}dx}[/tex]

is Riemann integrable, but not Lebesgue integrable. Note that this can only happen if the range is infinite. If the range is finite, then Lebesgue integrability is much stronger than Riemman integrability.
 
  • #9
disregardthat
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If I'm not mistaken, the integral

[tex]\int_0^{+\infty}{\frac{\sin(x)}{x}dx}[/tex]

is Riemann integrable, but not Lebesgue integrable. Note that this can only happen if the range is infinite. If the range is finite, then Lebesgue integrability is much stronger than Riemman integrability.
There really is no such thing as a riemann integral over an infinite interval. This integral is defined as the limit of riemann integrals, each of which can be considered lebesgue-integrals. Asking for the lebesgue-integral of that function on the [0,infinity) is simply not asking for the same thing.

That lebesgue-integrability is strictly stronger than riemann-integrability only applies to measurable functions on finite interval-domains. I don't know if there exists non-measurable riemann-integrable functions on finite intervals. I would doubt it though, but it's an interesting question.
 
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  • #10


If I'm not mistaken, the integral

[tex]\int_0^{+\infty}{\frac{\sin(x)}{x}dx}[/tex]

is Riemann integrable, but not Lebesgue integrable. Note that this can only happen if the range is infinite. If the range is finite, then Lebesgue integrability is much stronger than Riemman integrability.
Yes, there are examples of functions that are Riemann integrable but not Lebesgue integrable but my question is about functions that are not Riemann integrable.
 
  • #11
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Just noting that I've edited an earlier post of mine (with a lot of extra info), I hope it helps
 
  • #12
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There really is no such thing as a riemann integral over an infinite interval. This integral is defined as the limit of riemann integrals, each of which can be considered lebesgue-integrals. Asking for the lebesgue-integral of that function on the [0,infinity) is simply not asking for the same thing.

That lebesgue-integrability is strictly stronger than riemann-integrability only applies to measurable functions on finite interval-domains. I don't know if there exists non-measurable riemann-integrable functions on finite intervals. I would doubt it though, but it's an interesting question.
Well, to my knowledge, Riemann integrals over an infinite interval are still called Riemann integrals. They've done this is a lot of books I've read.

And indeed: if a function is non-measurable, then it cannot be Riemann-integrable. And with measurable, I mean Lebesgue-measurable. There exist functions which are not Borel-measurable, but which are Riemann-integrable. But it is true that Riemann-integrable functions are Lebesgue-measurable, the proof can be found in standard measure theory texts...
 
  • #13
disregardthat
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As you can verify, the sequence [tex]f \chi_{\epsilon_n}[/tex] with [tex]\epsilon_n := 1/n[/tex], fullfills the requirements of the theorem, such that [tex]\int_{Lebesgue} f(x) \mathrm dx = \int_{Lebesgue} \lim_n f(x) \chi_{\epsilon_n} \mathrm d x = \lim_n \int_{Lebesgue} f(x) \chi_{\epsilon_n} \mathrm d x = \lim_n \int_{Riemann} f(x) \chi_{\epsilon_n} \mathrm d x = \lim_n \ln{1} - \ln{\epsilon} [/tex]
What is happening here in the last line? This integral (inside the limit operator) is 0, not [tex]- \ln{\epsilon_n}[/tex] (as I assume you meant).
 
  • #14
disregardthat
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Well, to my knowledge, Riemann integrals over an infinite interval are still called Riemann integrals. They've done this is a lot of books I've read.
Perhaps, but not to my experience. It's not very important, but I doubt that the original definition would coincide with this extension of the term.

There exist functions which are not Borel-measurable, but which are Riemann-integrable.
Oh, really? I would really like to see this. That was what I meant by the way, my point was that lebesgue-integrability is stronger than riemann-integrability only for lebesgue-measurable functions.
 
  • #15
disregardthat
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the Lebesgue-integral (over the whole real line) of [tex]\sin{x}/x[/tex] really is not defined). The thing is: f is a positive measurable function, and for such functions we can always write down the integral (it's a theorem), the only thing is, the integral might be infinite (as was the case here).
Then extend it to 1 for x = 0! You are really not addressing the right point here. And even if one didn't make such an extension, the set where it isn't defined is a lebesgue null-set, so it is conventional that the integral is taken over the complement of this null-set.

EDIT II: this proof also answers another question of yours: one of the main methods of evaluating a lebesgue integral is writing it as a limiting case of Riemann-integrals (by use of limit-switching theorems)
This is as already established not true for all measurable functions. It will be for non-negative ones however.
 
  • #16
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Oh, really? I would really like to see this. That was what I meant by the way, my point was that lebesgue-integrability is stronger than riemann-integrability only for lebesgue-measurable functions.
Take a Lebesgue-measurable set which is not Borel-measurable. One can prove that you can find such a set inside the Cantor set. Then the characteristic function of this set is not Borel-measurable, but I think that it is Riemann-integrable. Indeed, we have a theorem that a function is Riemann-integrable if it's set of discontinuities has (Lebesgue-)measure 0. I think something like that should do the trick...
 
  • #17
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Jarle, I don't seem to be understanding any of the comments you've made to me, but let me try and reply the best I can:

concerning the limit of the integral: why is it zero? The integral of 1/x from epsilon to 1 is ln(1) - ln(epsilon), is it not?

"Then extend it to 1 for x = 0!" That was not all the point, I had even assumed one would ignore that point, it is irrelevant. The thing making it non-Lebesgue integrable is the fact that while the limit of that integral is finite, the integral of the absolute value is infinite, but I didn't want to get into so much detail, lest I confused the OP (as I had already put so much information in one little post)

"This is as already established not true for all measurable functions. It will be for non-negative ones however. "
I don't even understand what you're saying, on the level of English itself, but that might be because I'm foreign. Are you denying that limits of Riemann integrals are used for evaluating Lebesgue integrals?
 
  • #18
disregardthat
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Jarle, I don't seem to be understanding any of the comments you've made to me, but let me try and reply the best I can:

concerning the limit of the integral: why is it zero? The integral of 1/x from epsilon to 1 is ln(1) - ln(epsilon), is it not?
The function you were integrating was [tex]f(x) \chi_{\epsilon_n}[/tex], and this is non-zero only at a finite set.
 
  • #19
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You call [1/n; 1] a finite set?
 
  • #20
disregardthat
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"This is as already established not true for all measurable functions. It will be for non-negative ones however. "
I don't even understand what you're saying, on the level of English itself, but that might be because I'm foreign. Are you denying that limits of Riemann integrals are used for evaluating Lebesgue integrals?
Yes, for functions f(x) which take both positive and negative values it might be the case that the limit of riemann-integrals [tex]\lim_{r \to \infty} \int^r_0 f(x) dx[/tex] exists, but that f(x) is not lebesgue-integrable (over [tex][0,\infty)[/tex]). The example [tex]f(x) = sin(x)/x[/tex] verifies this.
 
  • #21
disregardthat
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You call [1/n; 1] a finite set?
It seems that you are not talking about the same function you were commenting on. He was considering f(x) = 1 for x = 1/N for integers N, and 0 elsewhere. I assumed you were talking of the same function. I have misunderstood your post.
 
  • #22
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"Yes, for functions..."

That's not denying me, that's just saying you can't always do it.
I was simply answering the OP's question about how to evaluate Lebesgue-integrable functions, and one of the main methods is, if possible, writing it as the limit of Riemann-integrals.

EDIT:
It seems that you are not talking about the same function you were commenting on. He was considering f(x) = 1 for x = 1/N for integers N, and 0 elsewhere. I assumed you were talking of the same function!
please re-read my post, I clearly stated that my edit was directed to the OP's last post, separate from the earlier issue. Also, I clearly defined the functions I used in the EDIT, so please read carefully before you reply, because now we're almost throwing a spam fest on a perfectly valid thread
 
  • #23
2,111
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No. For example
f(x) = 1 when x = 1/N for any integer N
f(x) = 0 otherwise.

f(x) is not Riemann integrable but it is Lebesgue integrable.
A bounded function on a compact interval is Riemann integrable if and only if it is continuous almost everywhere.
These two statements are contradictory, because the defined f(x) is continuous almost everywhere.
 
  • #24
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Indeed, so I believe Aleph's function gives Riemann integral zero
 
  • #25


Just noting that I've edited an earlier post of mine (with a lot of extra info), I hope it helps
your post was extremely enlightening, thank you. although i kind of lost jarle and you in posts following :-)
 

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