# Is the Riemann integral of a strictly positive function positive?

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

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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
The second link in this question doesn't seem to work. –  Jonas Meyer Jan 26 at 22:39

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 $\displaystyle \int_I f \geq \int_K f \geq \alpha \cdot \mu(K)>0$.

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@julien But Riemann integrable $\implies$ Lebesgue integrable, hence how could a counterexample exist? –  Did Apr 4 '13 at 15:13
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.

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