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?


8 Answers 8


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.

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    $\begingroup$ 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. $\endgroup$
    – Rachel
    Commented Mar 18, 2011 at 14:24
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    $\begingroup$ 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]. $\endgroup$
    – KCd
    Commented Mar 20, 2011 at 19:40
  • $\begingroup$ @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. $\endgroup$
    – Rachel
    Commented Mar 20, 2011 at 22:45
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    $\begingroup$ Its worth mentioning that the Henstock Kurzweil integral which is only a mild generalization of Riemann integration IS defined on intervals like (0,∞) though, and equality between that definition and the improper limit becomes a theorem. If the improper limit converges, you can still approximate arbitrarily well with Riemann sums without taking one limit first and then the other, but the summation points have to spread out as they go to infinity, which is the only reason why the naive constant-delta definition fails. $\endgroup$
    – saolof
    Commented Oct 2, 2021 at 9:22

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.

  • $\begingroup$ 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. $\endgroup$
    – KCd
    Commented Mar 20, 2011 at 19:38
  • $\begingroup$ @KCd: yes, I'll add the measurability requirement. $\endgroup$ Commented Mar 20, 2011 at 21:49
  • $\begingroup$ I would also add (heh) that part of the point of Lebesgue integration is to simply add up the areas with no regard to how we've added them up. On the other hand, Riemann integrals inherently have a direction to them which can create---in a sense---artificial cancellation with improper integrals. This is a bit close to Riemann rearrangement theorem type arguments regarding conditional convergence. We have made a sort of regularization choice with improper Riemann integrals and taking different approaches may lead to any real number we want. $\endgroup$ Commented Nov 25, 2021 at 17:18

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.

  • $\begingroup$ 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. $\endgroup$ Commented Mar 18, 2011 at 19:00
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    $\begingroup$ 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) $\endgroup$
    – user8268
    Commented Mar 18, 2011 at 20:05

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 evaluating 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 $0$.)


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.


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.


If one thinks about an integral as giving you the area under a curve, this area should be expected to behave in a natural fashion, ie, if you chop it up into pieces and rearrange them via translations, the area should be preserved.

This property will hold for any $L^1$ function, but it need not be true for a function which has only an improper integral.

It might be helpful to think about infinite sums- absolutely convergent sums can be thought of as integrals of $L^1$ functions. Sums which are convergent but NOT absolutely convergent have peculiar properties. For example, take $\displaystyle \sum_{n=1}^{\infty} \frac{ (-1)^n }{n} $.

Although this sum has a classical limit, we may permute the terms in the sum to allow the limit to take any finite value. As a rough sketch of this process, we can consider the positive elements and the negative elements separately — each of these sums must diverge on their own, or the sum would converge absolutely.

Given any target sum $S$, we can take positive terms until the partial sum is greater than $S$, and then add negative terms until the partial sum is below $S$. Repeating in this fashion, the partial sums will oscillate around $S$, with the error term shrinking since the positive and negative terms tend to zero by convergence of the original sum.

Clearly there is something funny going on here: We can get a number for the sum of the series, but by rearranging we could get different numbers. In this same fashion we could construct functions with the same property — if we chop up the graph of the function and rearrange it, we can change the volume of the graph. This means that we can't really assign a meaningful notion of volume to the graph of a function if it isn't integrable in absolute value.

  • $\begingroup$ but people cstill sum the series. in fact even more methods of summation. $\endgroup$ Commented Apr 16, 2020 at 12:50
  • $\begingroup$ IN fourier analysis throught one needs improper integrals. they are definitely not as funny as condtionally convergent series. $\endgroup$ Commented Apr 16, 2020 at 12:51
  • $\begingroup$ Further in cauchy residue theorem you find cauchy principal values needed often in number theory physics. $\endgroup$ Commented Apr 16, 2020 at 12:52
  • $\begingroup$ so improper integrals exist and nned to be studied. $\endgroup$ Commented Apr 16, 2020 at 12:53
  • $\begingroup$ Henstock kurzweil integral allows one to consider many improper integrals as it natural. $\endgroup$ Commented Apr 16, 2020 at 12:53

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.


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