Elementary proof of $f>0$ implies $\int f>0$? The question (Abbott, Understanding Analysis 2ed, 7.4.4) is: 
Show that if $f(x)>0$ for all $x\in[a,b]$ and $f$ is integrable, then $\int_a^b f>0$.
I can show it using Baire's theorem (the sets $E_n=\{x: f(x)>1/n\}$ can't all be nowhere dense...), but that's optional in this book, and the Lebesgue characterization of integrable functions is two sections ahead.  Is there a way using not much more than the definition of Riemann integral?
 A: The statement is true because one can show there exists $\xi \in (a,b)$ such that $$\int_a^b f(x) \, dx \geq f(\xi) (b-a)$$
In fact assume that $$f(x) > \frac 1{b-a} \int_a^b f(y)\,dy$$ for all $x \in (a,b)$
Changing the values of $f$ at $a$ and $b$ if necessary (which doesn't alter the value of the integral), we have a new function $\hat{f}$ such that $$\hat{f}(x) > \frac 1{b-a} \int_a^b \hat{f}(y)\,dy$$ for all $x \in [a,b]$
Remember now the theorem discussed in this question.
Since $\hat{f}$ is continuous at some $c \in (a,b)$, we can find $\varepsilon > 0\,$ so that $$\hat{f}(x) > \frac 1{b-a} \int_a^b \hat{f}(y)\,dy + \varepsilon$$ for all $x \in (c-\varepsilon,c+ \varepsilon) \subset (a,b)$.
Then, if we consider the partition $P=\{a,c-\varepsilon,c+\varepsilon,b\}$, we obtain $$\int_a^b \hat{f}(x)\,dx \geq L(\hat{f},P) \geq \int_a^b \hat{f}(x)\,dx + 2\varepsilon^2 > \int_a^b \hat{f}(x)\,dx$$ which is absurd.
The paper Rodrigo Lopez Pouso, Mean Value Integral Inequalities , Real Anal. Exchange Volume 37, Number 2, (2011), 439-450 is worth reading not only as the source of this proof.
A: As a sketch, try to write a Riemann-type sum with terms $f(x^*_j) \Delta x_j$ and notice that all terms in the sum are positive. Go by comparison.
