If I have the measure space $(\mathbb{R},\mathcal{B},\lambda)$ with the Lebesgue-measure $\lambda$ and the improper Riemann-integral of $\lvert f\rvert$ does not exists. Is then $f$ not Lebesgue-integrable?

For example I found here:

$\int_{\mathbb{R}}\frac{1}{1+\lvert x\rvert}\, d\lambda$.

It is $\lim_{R\to\infty}\int_0^R \frac{1}{1+\lvert x\rvert}\, dx=\infty$.

Does this mean that the function $\frac{1}{1+\lvert x\rvert}$ is NOT Lebesgue-integrable, i.e. that $\int_{\mathbb{R}}\left\lvert\frac{1}{1+\lvert x\rvert}\right\rvert\, d\mu=\int_{\mathbb{R}}\frac{1}{1+\lvert x\rvert}\, d\mu\geq\infty$?

I only know the fact, that, if the improper Riemann-integral of $\lvert f\rvert$ exists, that then $f$ is Lebesgue-integrable...

  • $\begingroup$ I fail to see the connection between your question and your example. But yes: if $|f|$ is a function that is continuous almost everywhere whose improper Riemann integral doesn't exist, then $f$ is not Lebesgue-integrable. Your example has a value neither under Riemann integration nor Lebesgue integration. $\endgroup$ – Omnomnomnom Dec 12 '13 at 18:11
  • $\begingroup$ It is possible that the function may be Lebesgue integrable but that the Riemann integral does not exist, for example, $f=1_\mathbb{Q}$. $\endgroup$ – copper.hat Dec 12 '13 at 18:14

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.