7
$\begingroup$

I wonder why we say that $f$ is integrable iff $\int|f|\,d\mu$ is finite? Why we use absolute value? Won't it be enough to have that $\int f\, d\mu$ is finite to call $f$ integrable? Are there functions $f$ for which $\int f \ d\mu$ is finite but $\int |f| \ d\mu$ is not (like conditional convergence for series)?. But then why we would call such $f$ not Lebesgue integrable?

$\endgroup$
  • 4
    $\begingroup$ As with sequences, absolute convergence frees you from having to think of a particular order of summation. Loosely, the power of Lebesgue integration is that it 'frees' you from a particular slicing of the domain of integration. $\endgroup$ – copper.hat Jan 7 '15 at 20:19
  • 2
    $\begingroup$ Well, the answer is because that is the definition of integrability in a Lebesgue sense. I am trying to convey what absolute summability brings to the party. $\endgroup$ – copper.hat Jan 7 '15 at 20:22
  • 1
    $\begingroup$ consider $f(x)=-1$ for $x<0$, $f(x)=1$ for $x\geq 0$. is $\int f<\infty$? or $\int\sin x/x$, etc. $\endgroup$ – yoyo Jan 7 '15 at 20:26
  • 2
    $\begingroup$ If a sequence is absolutely summable, then you can rearrange the summation anyway you want. This is not true with summable but not absolutely summable sequences. For example, you can rearrange the sequence formed from $(-1)^n {1 \over n}$ to get any number you want. $\endgroup$ – copper.hat Jan 7 '15 at 20:26
  • 1
    $\begingroup$ Think to the $\operatorname{sinc}$ function: it is Riemann integrable on $\mathbb{R}^+$ since $\lim_{t\to +\infty}\int_{0}^{t}\operatorname{sinc}(x)\,dx$ exists, but if we integrate it on a set $E$ that is concentrated around $\frac{\pi}{2}+2\pi\mathbb{Z}$, we can make the Lebesgue integral over $E$ arbitrarily large. In order to avoid such issues, dealing with non-negative measures requires to deal with non-negative functions, so we say that $f\in L^1$ if $\int |f|\,d\mu$ is finite. $\endgroup$ – Jack D'Aurizio Jan 7 '15 at 20:27
2
$\begingroup$

It's very much like conditional convergence for series. A conditionally convergent series (i.e. a convergent series whose partial sums of the positive terms diverge to $+\infty$ and those of the negative terms to $-\infty$) converges to a different thing if you suitably rearrange its terms.

Consider the integrals $$ \int_0^1\left(\int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2}\, dy\right)\,dx \quad\text{and} \quad \int_0^1 \left(\int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2}\, dx\right)\,dy. $$ You can show by freshman calculus methods that one of these is $\pi/2$ and the other is $-\pi/2$. Rearranging it has changed the sum.

A simpler example is $$ \lim_{a,b\to\infty\text{ ???}} \int_b^a \frac{x\,dx}{1+x^2}. $$ If you evaluate $$ \lim_{a\to\infty}\int_{-a}^a \frac{x\,dx}{1+x^2} $$ you get $0$, but $$ \lim_{a\to\infty}\int_{-a}^{2a}\frac{x\,dx}{1+x^2} $$ is a non-zero number. Try it and see.

| cite | improve this answer | |
$\endgroup$
9
$\begingroup$

As far as why we define integrability this way, one reason that I can think of is that we cannot always assign a numerical value to $\int f d\mu$, but with $\int |f| d\mu$, we can - it's either finite or infinite. That is, $\int f d\mu$ is not always well-defined. However, if we know that $\int |f| d\mu < \infty$, then $\int f d\mu$ is well-defined since $$\int |f| d\mu = \int f^+ d\mu + \int f^- d\mu < \infty$$ which is true only when both positive and negative parts of the function are both finite. Thus, $$\int f d\mu = \int f^+ d\mu - \int f^- d\mu $$ is finite as well.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ But if integral of $f$ is well defined and is finite, can be then integral of $|f|$ infinite? (like with ahrmonic series) $\endgroup$ – luka5z Jan 7 '15 at 20:46
  • 3
    $\begingroup$ No. The Lebesgue integral of a real-valued measurable function, $f = f^+ - f^-$, is defined to be $\int f d\mu = \int f^+ d\mu - \int f^- d\mu$, but only when one of those parts is finite. $\int f d\mu $ has no meaning if both parts are infinite. So if you insist that $\int f d\mu = \int f^+ d\mu - \int f^- d\mu < \infty$, then both parts have to be finite, since at least one of them has to be in order to even be defined. But this means that $\int |f| d\mu = \int f^+ d\mu + \int f^- d\mu <\infty$ too. $\endgroup$ – dannum Jan 7 '15 at 21:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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