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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?

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  • $\begingroup$ BTW, you need to assume something like $f\geq 0$. $\endgroup$ – Stefan Lafon Aug 7 '19 at 22:57
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    $\begingroup$ $f$ is given to be "nonnegative" in the first sentence. $\endgroup$ – Bladewood Aug 7 '19 at 23:03
  • $\begingroup$ Ah sorry, I was going by the title. $\endgroup$ – Stefan Lafon Aug 7 '19 at 23:05
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    $\begingroup$ 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 ? $\endgroup$ – Matematleta Aug 7 '19 at 23:16
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    $\begingroup$ No such thing as Riemann Lebesgue Theorm is required. The question involves measure theory and its answer is quite trivial. $\endgroup$ – Kavi Rama Murthy Aug 7 '19 at 23:25
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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.

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