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.

In the proof here a strictly positive function in $(0,\pi)$ is integrated over this interval and the integral is claimed as a positive number. It seems intuitively obvious as the area enclosed by a continuous function's graph lying entirely above the x-axis and the x-axis should not be zero. But how can I prove this formally?

If the function is positive over a closed interval apparently the result is not true. This has further confused me. Can someone please clarify my doubt.

Thanks

share|improve this question
    
If you have a counterexample function $f$ s.t. $f(x)>0$ on the closed interval $[a,b]$ but $\int_a^b f = 0$, then clearly $f(x)>0$ on $(a,b)$ and $\int_a^b f = 0$. –  Ilya Apr 4 '13 at 14:53
    
Any graph argument implicitly assumes that the function is continuous. In which case $f>0$ implies $\int_a^b f>0$. –  1015 Apr 4 '13 at 14:53
    
to add to a comment by @julien, although Riemann integrability does not require continuity, the function in your first link is continuous –  Ilya Apr 4 '13 at 14:56
    
If $f$ is Riemann integrable, then $f$ is continuous a.e., and one just needs continuity at a single point to show $\int_a^b f > 0$. –  copper.hat Apr 4 '13 at 16:10

2 Answers 2

up vote 3 down vote accepted

Let $f : I \to \mathbb{R}$ be a function on some interval $I$.

If $f$ is continuous and positive on $I$, then $\displaystyle \int_I f >0$. Indeed, $f \geq \alpha >0$ on some closed interval $K \subset I$, so $\displaystyle \int_I f \geq \int_K f \geq \alpha \cdot \mu(K)>0$.

share|improve this answer
    
@julien But Riemann integrable $\implies$ Lebesgue integrable, hence how could a counterexample exist? –  Did Apr 4 '13 at 15:13
1  
A Riemann integrable function $f$ over a nondegenerate interval has points of continuity. This is enough to insure the integral is positive for positive $f$. –  David Mitra Apr 4 '13 at 15:15
    
@DavidMitra I was impressed by Remark 4.21 here. And I did not really think about it any further... –  1015 Apr 4 '13 at 15:19
    
@Did Sure, I was silly. Do you have any idea of what they mean here in Remark 4.21? –  1015 Apr 4 '13 at 15:29
    
@julien: I can only imagine that they mean with the machinery developed so far, they cannot prove it. However, I think it is straightforward to prove, let me see if I can dig up a proof. –  copper.hat Apr 4 '13 at 15:41

If $f$ is non-negative, Riemann integrable and $\int_a^b f(x) dx = 0$ (with $a<b$) then it must be the case that $f(x) = 0$ a.e. Hence if $f$ is strictly positive on $[a,b]$, then it must be the case that $\int_a^b f(x) dx > 0$.

This is straightforward to see using the Lebesgue integral.

See Corollary 3 in www.math.sc.edu/~schep/riemann.pdf for a straightforward proof. The essence is to show that for all $c > 0$ the set $\{x \in [a, b] | f (x) \ge c \}$ has content zero.

share|improve this answer

Your Answer

 
discard

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.