2
$\begingroup$

Folland defines the integral of a non-negative measurable function $f$ to be $$\int f d\mu = \sup \left\{ \int \phi d\mu : 0 \leq \phi \leq f, \phi \text{ simple}\right\}$$

after defining the integral of a non-negative simple function. By this definition, we should have that, for any measurable set $E$: $$\int_{E} f d\mu = \int f \chi_{E} d\mu$$

But isn't this definition wrong? I think it works fine if we require $f \in L^{1}$, but this definition contradicts usual definitions we use when we talk about convergence. Namely, consider the following:

$$\int_{\{0\}} \frac{dx}{|x|} = \int \chi_{\{0\}} \frac{dx}{|x|}=0$$

and compare it with the definition:

$$\lim_{\epsilon \to 0} \int_{[-\epsilon, \epsilon]} \frac{dx}{|x|} = \infty$$

Do I have a point here?

$\endgroup$
  • 6
    $\begingroup$ The limit you've written has nothing to do with the definition of an integral $\endgroup$ – Omnomnomnom Dec 8 '14 at 1:43
  • $\begingroup$ I don't mean to imply any relation of the limit definition with Folland's, but it was more my point that his definition makes it unclear how to interpret $\int_{\{0\}} \frac{dx}{|x|} $. Sure, the limit I gave isn't using his definition of integral, but is it not a valid way to define such an integral? I guess it's only problematic to the extent that it isn't consistent with other meanings we assign to such an integral. And this isn't really something unavoidable in defining the Lebesgue integral; I believe the definition Stein and Royden use are consistent with my limit definition. $\endgroup$ – christoph Dec 8 '14 at 1:53
  • $\begingroup$ Your limit is a valid extension of the Riemann integral of the function. However, it doesn't work here. $\endgroup$ – Omnomnomnom Dec 8 '14 at 2:00
  • $\begingroup$ I wonder how can we require $f\in L^1$ before defining what is an integral of $f$. $\endgroup$ – Ilya Dec 8 '14 at 10:33
  • $\begingroup$ I know what that sounds like, but I wasn't trying to redefine Folland's integral. I was just trying to point out that, if that were the case, then there wouldn't be ambiguity with definition the integral at the singleton $\{0\}$. $\endgroup$ – christoph Dec 8 '14 at 14:56

Your Answer

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

Browse other questions tagged or ask your own question.