Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The function $f(x) = \sin(x)/x$ is Riemann Integrable from $0$ to $\infty$, but it is not Lebesgue Integrable on that same interval. (Note, it is not absolutely Riemann Integrable.)

Why is it we restrict our definition of Lebesgue Integrability to absolutely integrable? Wouldn't it be better to extend our definition to include ALL cases where Riemann Integrability holds, and use the current definition as a corollary for when the improper integral is absolutely integrable?

share|improve this question

7 Answers 7

up vote 25 down vote accepted

You could just as well ask the opposite question: why do we define Riemann integration in such a way that an integral can be convergent without being absolutely convergent? The definition of each type of integral "is what it is," and the way the Lebesgue definition is defined there is no need for improper integrals like in Riemann integration.

We could simulate an improper integral with Lebesgue integration by taking a limit of Lebesgue integrals over bounded regions. But that's not something that's usually of interest in the Lebesgue theory.

The things that are of interest are convergence theorems like the dominated convergence theorem.

Dominated convergence theorem: If $(f_n)$ is sequence of measurable functions, $|f_n|<|g|$ for $n \in \mathbb{N}$, $\int |g| < \infty$, and $f_n\to f$ pointwise then $\int f < \infty$ and $\int f_n \to \int f$.

In that theorem, the dominating function $g$ needs to be absolutely integrable. Your example, in fact, can be modified to give a counterexample to this statement:

False: if $(f_n)$ is sequence of measurable functions, $|f_n|<|g|$ for $n \in \mathbb{N}$, $\int g < \infty$ when the integral is computed in the improper sense, and $f_n\to f$ pointwise then $\int f < \infty$ (again in the improper sense) and $\int f_n \to \int f$.

The actual theorem requires $\int |g|$ to be finite. Since that is the sort of condition that we work with most of the time, we use the word "integrable" for it to save space. We can still recapture improper integrals if we have to, but they're rarely of interest in the context of Lebesgue integration, so we don't want to spend a good word like "integrable" on them.

share|improve this answer
    
You ought to include hypotheses about your functions f_n in the dominated convergence theorem. It is not a theorem about arbitrary (e.g., nonmeasurable) pointwise convergent sequences of functions that are bounded in absolute value by a Lebesgue integrable function. –  KCd Mar 20 '11 at 19:38
    
@KCd: yes, I'll add the measurability requirement. –  Carl Mummert Mar 20 '11 at 21:49

Technically speaking, the function $\displaystyle{f(x) = \frac{\sin x}{x}}$ is not Riemann integrable on $(0, \infty)$, but rather improperly Riemann integrable on $(0, \infty)$.

The construction of the Riemann integral only works for bounded intervals. We can extend this construction to unbounded intervals like $(0, \infty)$, but that requires an additional limiting process. It is the first construction (Riemann integrals for bounded integrals) that the Lebesgue integral generalizes.

share|improve this answer
6  
Actually, this helped clarify things for me. I was not overtly aware that the Riemann integral was constructed for bounded intervals only, and that the Lebesgue integral generalizes THIS notion. Moreover, as @Carl states in his answer, the Lebesgue integral can be extended much the same way the Riemann integral can on an unbounded interval. –  Rachel Mar 18 '11 at 14:24
1  
Rachel, if you were not overtly aware that the Riemann integral is not directly defined on unbounded intervals, think about its definition in terms of Riemann sums: that requires a bounded interval [a,b]. –  KCd Mar 20 '11 at 19:40
    
@KCd - Thanks for the comment. It actually prompted me to go back and look through the section on Riemann integration in Rudin's little analysis book. He barely mentions improper integration, and leaves it as an exercise! Otherwise, everything is confined to treatments on bounded intervals. –  Rachel Mar 20 '11 at 22:45

You might be interested in Henstock-Kurzweil (HK) integral. Its definition is an easy modification of Riemann integral. All Lebesgue-integrable functions are integrable in the HK-sense, and so is your function $\sin(x)/x$. The usual theorems from Lebesgue theory (such as the dominated convergence theorem) have extensions to the HK theory. On top of that, Newton-Leibniz formula holds for any function admitting an anti-derivative (which is not true in Lebesgue theory).

That being said, you definitely need Lebesgue theory for spaces different from $\mathbb{R}^n$. Notice also that $\mathbb{R}^2$ is isomorphic to $\mathbb{R}$ as a measure space (and that any sensible measure space is isomorphic to an interval); you would lose this unity if you wanted e.g. $\sin(x)/x$ to be integrable.

share|improve this answer
    
It's true that there is an extension of the dominated convergence theorem (DCT) in the H-K theory, but it can't apply to integrals like $\int_0^\infty \sin(x)/x$, because that function trivially dominates its positive part although the positive part has an infinite integral. So there has to be some positivity requirement on the DCT even in the H-K theory; I'm not sure how they usually state the DCT there. –  Carl Mummert Mar 18 '11 at 19:00
1  
If I remember correctly, in the HK version of DCT you demand a bound $g<f_n<h$ with $g$ and $h$ HK-integrable. (This implies that $h-g$ is HK-integrable, and hence (as $h-g\geq 0$) Lebesgue integrable, so it's not a big extension of the Lebesgue case; the HK formulation seems quite natural though) –  user8268 Mar 18 '11 at 20:05

I'll add an additional answer based on a response from my measure theory professor, as it may be of use.

Essentially, his response was that the purpose of Lebesgue Integration is to make the set of integrable functions complete. (Recall that we can form a sequence of Riemann Integrable functions that converge to a function that is not Riemann Integrable.) As was noted by @Carl Mummert in his answer, the things of interest in Lebesgue Theory are convergence theorems (which tie into this notion of completeness), not a theory of integrals for specific classes of functions.

As such, it isn't that the definition of Lebesgue Integrability isn't as broad as it might be, but that it is broad enough to ensure completeness.

share|improve this answer

No one seems to have mentioned Fubini's theorem yet. Consider this example: $$ \int_0^1 \int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} \,dy\,dx = \frac\pi4, $$ but $$ \int_0^1 \int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} \,dx\,dy = -\frac\pi4. $$ Simply interchanging $dy$ and $dx$ changes the value of the integral. Fubini's theorem says that can't happen if the integral of the absolute value is finite.

So that's another reason why absolute convergence is important.

(Without evaluting either of the two integrals above, one can quickly see that interchanging $dx$ and $dy$ is the same as leaving them as they are but interchanging $x$ and $y$ elsewhere, and in this case that's merely reversing the order of subtraction; hence multiplying the bottom line by $-1$. Thus it can remain the same only if the integral is $0$.)

share|improve this answer

Often, e.g. in probability theory, one wants to integrate over a space quite different from $\mathbf{R}^n$ where there's no clear standard way to find a limit as the space over which one integrates approaches the whole measure space.

share|improve this answer

Yes, absolute convergence is important. The strength of the Lebesgue integral lies in convergence theorems but yes we lose the simplicity by having improper integrals not Lebesgue. A satisfactory theory is given by the Henstock-Kurzweil integral. Moreover every derivative is integrable in that theory and this is a shortcoming of Lebesgue theory that it does not allow to do so. Also we cannot have mean value thorem for Banach space valued mappings in integral form unless we use Henstock-Kurzweil. Nothing prohibits from restricting to the subfamily of absolutely integrable mappings when there is need to do so.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.