Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I was a bit confused by this link mentioned in this question - in particular, in Remark 4.21:

Suppose that $f$ is a positive function on $[a,b]$. If $f$ is Henstock-Kurzweil integrable, then the best we can conclude is that $\int_a^bf\geq 0$; from our results so far we cannot conclude that the integral is positive. The Riemann integral has the same defect.

The latter part of the statement is something that is not very clear to me. For any non-negative Lebesgue measurable function $f:[a,b]\to\Bbb R_+$ it holds that if $\lambda\{f>0\}>0$ then $$ \int_{[a,b]} f\mathrm d\lambda >0 $$ where $\lambda$ is the Lebesgue measure and the integral is the Lebesgue one. As a result, if there is a function $f:[a,b]\to\Bbb R_+$ such that $f(x)>0$ for any $x\in [a,b]$, but $\int_a^bf = 0$ (whatever way the latter integral is defined) then $f$ is either non-measurable, or the latter integral is such that $$ \int_a^b f\neq \int_{[a,b]}f\mathrm d\lambda. $$ So my question is: is there a positive function on $[a,b]$ such that $\int_a^b f = 0$ for some (commonly known) definition of integral. In particular, does Remark 4.21 above mean that such example exists for the Riemann integral?

share|improve this question
    
perhaps when they say "positive" they mean "non-negative". Otherwise, I am as confused as you. –  robjohn Apr 5 '13 at 8:04
    
Not that I know of. –  Kaster Apr 5 '13 at 8:07
    
@robjohn: perhaps, but is there a need to say that about non-negative function? Clearly, $f = 0$ is non-negative and has zero integral. Also, further in the Remark the show that the Lebesgue integral of a positive function is strictly positive. –  Ilya Apr 5 '13 at 8:16
    
The key passage here seems to be from our results so far. Once you know that the Lebesgue integral extends the Riemann integral (Theorem 3.103) the argument you give shows that there is no such example. I think that the authors try to emphasize at this point that the proof of this result needs more machinery for the Riemann integral while it is simple for the Lebesgue integral. –  Martin Apr 5 '13 at 14:46
    
@Martin: perhaps, that's indeed the point. –  Ilya Apr 5 '13 at 15:02

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.