1
$\begingroup$

Consider the following statement:

Let $I$ be a bounded open interval and let $f$ be a real function that is continuous on $I$. Then, $f$ is integrable on $I$

This statement is false, but I cannot understand why. I thought that a continuous function on a bounded interval will be bounded, which seems to imply integrability.

Edit: Also, why is it then that $\lvert\int f(x)dx\rvert$ from $a$ to $b$, where $a,b$ are in the interval $I$, is strictly bounded?

$\endgroup$
  • $\begingroup$ A continuous function on a bounded closed interval must be bounded. $\endgroup$ – David C. Ullrich Jan 1 '16 at 19:19
  • $\begingroup$ Thank you! But why does this fail over an open interval? $\endgroup$ – john melon Jan 1 '16 at 19:19
  • $\begingroup$ Consider $f(x)=1/x$ on $x\in (0,1)$. It is continuous and non integrable. $\endgroup$ – Giuseppe Negro Jan 1 '16 at 19:20
  • $\begingroup$ Ah! I see. But is the core reason why $1/x$ is not integrable over $(0,1)$ the fact that it is unbounded? $\endgroup$ – john melon Jan 1 '16 at 19:21
  • $\begingroup$ Are you working in the sense of Lebesgue, proper Riemann, or possibly-improper Riemann? (In any case, bounded "reasonable" functions are integrable, where "reasonable" has different meanings in each case). $\endgroup$ – Ian Jan 1 '16 at 19:33

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.