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.
Best Answer
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.