2
$\begingroup$

Let $E \subseteq \mathbb{R}^n$ be a open subset and let $f:E \to \mathbb{R}$ be positive (not nessesarily bounded) and a.e. continuous and suppose that the Lebesgue-integral of $f$ over $E$ exists (i.e. is finite). Is it true, that then $f$ is improper Riemann integrable over $E$?

The improper Riemann integral is defined as follows:

For every $M > 0$ let $g_M = \min\{f,M\}$. Then $f$ is improper Riemann integrable iff $\lim_{M \to \infty} \int_E g_M(x)dx$ exists, where the integral is the Riemann integral.

$\endgroup$
  • $\begingroup$ Yes, of course. Thank you $\endgroup$ – Andrei Kh May 11 '18 at 16:32
  • $\begingroup$ Is $E$ Jordan measurable? Notice there are open sets which aren't Jordan measurable. $\endgroup$ – user251257 May 11 '18 at 16:44
  • $\begingroup$ Yes, I forgot to mention it. But thanks for the counterexample in this case! $\endgroup$ – Andrei Kh May 11 '18 at 17:33
0
$\begingroup$

If $M_n\to \infty$ then $g_{M_n} \to f$ increasingly. So $\int g_{M_n} \to \int f$ (even if the latter is infinite). So for positive functions, $f$ is R integrable improperly if and only if $f$ is L integrable. You can do any other truncations for $f$ positive, provided they converge to $f$ increasingly.

$\endgroup$
0
$\begingroup$

If $E$ is Jordan measurable, then the statement is true due to one of many convergence theorem of Lebesgue integral.

If $E$ is not Jordan measurable, then it is wrong. Let $E$ be the complement of the Fat Cantor set in $(0,1)$. Then, $E$ is open and not Jordan measurable. In particular, it's indicator $\chi_E$ isn't Riemann integrable but Lebesgue integrable.

$\endgroup$

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.