Can one prove $\int f > 0$ for $f > 0$ without Lebesgue integration? In an exercise session of an analysis course (which covers Riemann integration and differentiation in one dimension rigorously) the following question came up:
Suppose $f$ is strictly positive and integrable (on some compact interval $[a,b]$ on the real line). Can we show that $\int_a^b f > 0$?
The proof by using measure theory and Lebesgue integration is easy, but also beyond the students at this point. So can one do without machinery such as sets of zero measure?

Tools in use: Mean value theorems, fundamental theorem of calculus, Riemann's condition for integrability (upper and lower integral within every epsilon of each other implies integrability), Riemann-Darboux integral, usual integration and differentiation techniques, as well as elementary real analysis in the epsilon-delta style.
 A: This can be shown using the concept of oscillation.
For a bounded $f$, the oscillation of $f$ over set $A$ (which is not a single point) is given by
$$w(A) = \sup_{A} f - \inf_{A} f$$ 
For a single point $x$, the oscillation is defined as
$$ w(x) = \inf_{J} w(J)$$
where $J$ ranges over bounded intervals containing $x$.
Note that if $x \in I$, then $w(x) \le w(I)$.
Now if $f$ is Riemann integrable over $[a,b]$, then we can show that given any $n \gt 0$, there is a sub-interval $I_{n}$ of $[a,b]$ such that $w(I_n) \le \frac{1}{n}$.
This is because, if every subinterval $I$ of $[a,b]$ had $w(I) \gt \frac{1}{n}$, then for every partition of $[a,b]$ the difference between the upper and lower sums would be at least $\frac{b-a}{n}$ and as a consequence, $f$ would not be integrable.
Now pick $I_{n+1} \subset I_{n}$ such that $w(I_{n+1}) \le \frac{1}{n+1}$.
By completeness there is a point $c$ such that $c \in I_n$.
Thus $w(c) \le w(I_n) \le \frac{1}{n}$ for all $n$. Hence $w(c) = 0$.
Now it can be show that $f$ is continuous at point $x$ if and only if $w(x) = 0$.
Note: This is basically a simplification of one proof of the Riemann Lebesgue theorem of continuity almost everywhere.
A: If f is integrable in [a,b], there is a point $x_0$ in the interval where f is continuous, and positive.
Then there is a neighborhood of the point where f is positive, contained in the interval.
There we can take a closed interval where the function is positive and there f has minimum then the integral at this interval is bigger or equal than the integral of the minimum, which is positive. We can then apply additivity in the interval [a,b] and in the other closed intervals. The integral is not negative since each Riemann sum is non-negative.
A: While the essential ideas here are already present in the existing answers, I'd like to present my version of this solution, first composed in 2005.
If the function is continuous at any point, the integral is clearly positive- you can stick a box under the graph. If we can show that a Riemann-integrable function is continuous somewhere, we will be done. There is a well-known theorem stating that the set of points of discontinuity has measure zero, but I'd like to stick with Riemann tools here.
Suppose $f$ is Riemann-integrable on the interval $[0,1]$. Choose a partition $P_1$ so that the lower sum and upper sum on $P$ differ by no more than 1. Then we can choose an interval of the partition $I_1$ with length $3\delta_1$ so that $\sup_{x\in I_1}f(x)-\inf_{y\in I_1}f(y)\le1$. Let $S_1$ be the closed middle third of $I_1$; then $|f(x)-f(x+t)|\le1$ for all $x\in S_1,|t|<\delta_1$.
Now choose a partition of $S_1$ with upper and lower sums differing by no more than $\frac12\delta_1$ and repeat the process. This leads to an interval $S_2$ of length $\delta_2$ with $|f(x)-f(x+t)|\le \frac12$ for all $x\in S_2,|t|<\delta_2$. Continuing, we find a nested sequence of closed intervals $S_n$ of length $\delta_n$ such that $|f(x)-f(x+t)|\le\frac1n$ for all $x\in S_n,|t|<\delta_n.$ This sequence has a nonempty intersection, and that point of intersection is a point of continuity.
This actually proves that the function is continuous at a dense set of points; we could have started with any subinterval.
