# 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?

• BTW, you need to assume something like $f\geq 0$. – Stefan Lafon Aug 7 '19 at 22:57
• $f$ is given to be "nonnegative" in the first sentence. – Bladewood Aug 7 '19 at 23:03
• Ah sorry, I was going by the title. – Stefan Lafon Aug 7 '19 at 23:05
• if you want to prove $f=0$ almost everywhere, then at some point you use Lebesgue theory. So what exactly do you mean by not using Lebesgue integral theory ? – Matematleta Aug 7 '19 at 23:16
• No such thing as Riemann Lebesgue Theorm is required. The question involves measure theory and its answer is quite trivial. – Kavi Rama Murthy Aug 7 '19 at 23:25

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.