# Is a differentiable function on $(-2, 4)$ always integrable on $[-2, 4]$?

So my question is, say I have a function that is differentiable on $(-2, 4)$. Is it always integrable on $[-2, 4]$?

I know that if $f$ is diff on $(-2, 4)$, then it is continuous on $(-2, 4)$. And I also know that if $f$ is continuous on $[-2, 4]$ then it is integrable on $[-2, 4]$. However, I am wondering if there is such a function so that there would be a problem at the endpoints of the closed interval so that it is differentiable on the open interval, but not integrable on the closed interval.

• Commented Nov 28, 2014 at 23:23
• There are a couple great books with titles like "famous counter-examples in mathematics" ... here are two such: smile.amazon.com/… and smile.amazon.com/… Commented Nov 29, 2014 at 19:14
• @Lucian: Could you explain why Volterra's function is of relevance? Commented Nov 30, 2014 at 11:30

The function $f(x)=\frac 1x$ is differentiable on $(0,1)$, yet it is not integrable on $[0,1]$.

edit

However, if you have a function $f$ which is differentiable on $[0,1]$, then it is necessarily continuous on $[0,1]$, hence measurable. Moreover, a continuous function on a compact is bounded, hence $f$ is bounded measurable, therefore integrable.

• This is a little bit cheating. Can you give an example where the intervals are both closed? Commented Nov 30, 2014 at 6:25
• @Mehrdad oops, my apologies, for some unknown reason I thought that you were the one who asked the initial question. Commented Nov 30, 2014 at 10:47
• @Mehrdad It is not cheating, it is merely an answer to the initial question=) As for closed intervals, see edit. Commented Nov 30, 2014 at 10:48
• Why don't you you answer the question with the given domain ? Commented May 21, 2015 at 11:17

Generally no, but with additional assumption that it is bounded -- yes. For the unbounded case we have easy counterexamples, as $1/x$ or $1/\ln x$ on $(0,1)$.

• It appears that the derivative of Volterra's function is bounded, but not (Riemann) integrable. Commented Nov 29, 2014 at 16:02
• @anorton Right, but its derivative is not differentiable. Commented Nov 29, 2014 at 17:28
• $\ln x$ is integrable on $[0,1]$, or maybe I misunderstand what you mean at the end there. Commented Nov 30, 2014 at 9:33
• @alex.jordan Oh, my misprint! $1/\ln x$. Corrected. Commented Nov 30, 2014 at 19:05