Elementary proof of $\int f=0\implies f=0$ a.e. for Riemann integrals Let $f:[a,b]\to\mathbb{R}$ be a nonnegative Riemann integrable function. I want to show that if $\int_a^bf=0$, then $f=0$ almost everywhere. I can think of two ways of showing this:


*

*Riemann integrable function is Lebesgue integrable, and the respective integrals are equal, so use the Lebesgue integral theory, in which the result is quite standard.

*According to Riemann-Lebesgue theorem, Riemann integrability implies almost everywhere continuity. Therefore one can show that $f=0$ wherever it is continuous, which is easy.


But is there a more elementary proof which does not use Lebesgue integral theory or big theorems like Riemann-Lebesgue theorem?
 A: Let's use the Darboux definition of the Riemann integral.
Let $\epsilon, \delta > 0$.  There must exist a partition $P = \{a = x_0 < x_1 < \dots < x_n = b\}$ for which the upper sum $U_{f,P}$ is less than $\epsilon \delta$.  Now each interval $[x_{i-1}, x_i]$ on which $f$ attains a value larger than $\delta$ contributes at least $\delta (x_i - x_{i-1})$ to the upper sum.  We conclude that the total length of those intervals is at most $\epsilon$.  In other words, the set $\{f > \delta\}$ is covered by a finite number of intervals with total length at most $\epsilon$, so its Lebesgue (outer) measure is at most $\epsilon$.  But $\epsilon$ was arbitrary, so $\{f > \delta\}$ has Lebesgue measure zero.
Now $\delta$ was also arbitrary, so taking $\delta = 1/k$, we have that $\{f > 0\} = \bigcup_k \{f > 1/k\}$ is a countable union of measure zero sets, hence has measure zero.  This needs only the countable subadditivity of Lebesgue measure which is elementary to prove.  Or to proceed more directly, fix $\eta > 0$ and use the previous construction with $\epsilon = \eta \cdot 2^{-k}$ to cover the set $\{f > 1/k\}$ by finitely many intervals of total length at most $\eta \cdot 2^{-k}$.  Unioning over $k$, we have a covering of $\{f > 0\}$ by countably many intervals of total length at most $\eta$, which by definition of Lebesgue outer measure means that $\{f > 0\}$ has (outer) measure at most $\eta$, and $\eta$ was arbitrary.
