0

I know that a function is integrable if : $$ \int f <\infty $$

but, what this say about $f$ ?

J. W. Tanner
  • 60,406
José Marín
  • 941
  • Add up a bunch of "spikes" where the $n$-th spike has height $n$ and width $1/n^3$. – Angina Seng Jun 24 '19 at 05:55
  • If $f$ changes sign it is possible that $\int f <\infty$ but $f$ is not integrable. What notion of integration are you referring to? – Conifold Jun 24 '19 at 05:57
  • @Conifold A Lebesgue Integral on $[0,1]$ – José Marín Jun 24 '19 at 06:00
  • 1
    Then $f$ is integrable iff it is measurable and $\int |f| <\infty$. It can be unbounded, $\int f <\infty$ can exist as improper integral even without $f$ being Lebesgue integrable. – Conifold Jun 24 '19 at 06:05
  • Suppose $f$ is continuous on a bounded interval $J$ and suppose $f(x)=0$ for $x\not \in J $. Suppose the classical (pre-Riemann ) integral $\int_Jf(x)dx $ exists. Then $that$ is the value of the Lebesgue integral $\int_{\Bbb R}f.$ For example $J=(0,1)$ and $f(x)=1/\sqrt x$ for $x\in J.$ – DanielWainfleet Jun 24 '19 at 07:43

1 Answers1

1

If you are talking about Lebesgue integrability, yes. For example, $f(x)=1/\sqrt{x}$ on $(0,1)$ is integrable. It is also integrable as an improper Riemann integral. But for a function on a segment $[a,b]$ to be Riemann integrable (in the proper sense) it has to be bounded.

GReyes
  • 16,446
  • 11
  • 16
  • So your defined function is not Riemann integral on $[0,1]$ but it is Lebesgue integrable on the same interval? – José Marín Jun 24 '19 at 06:02
  • 1
    Right. But the Riemann integral is extended to cover those cases (improper Riemann integral) and in that sense it is integrable as well. BTW there are functions that are integrable in the improper sense but are not in the Lebesgue sense (they must change sign wildly). – GReyes Jun 24 '19 at 06:06