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Suppose $f\geq 0$, is Riemann integrable on $[a,b]$, then $\int_a^bf=0$ iff $D=\{x\in[a,b]\mid f(x)>0\}$ has Lebesgue measure zero.

A set $A\subseteq\mathbb{R}$ has Lebesgue measure zero iff $\forall\epsilon>0\exists$a countable family of open intervals $\{I_n\}_{n\in\omega}(A\subseteq\bigcup_{n\in\omega}I_n\wedge\sum_{n=0}^\infty|I_n|<\epsilon)$.

So far my thoughts are, ($\rightarrow$) $L(P,f)=0$ for all partition $P$. Let $\epsilon>0$, then $\exists P_0(U(P_0,f)<\epsilon)$. But how can I construct a family of open intervals using that partition $P_0$?

I found some similar questions here and here, but they are not in the form I stated. I am just starting to learn measure theory. Can anyone prove it using definitions only? Thanks!

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Kaa1el
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1 Answers1

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We know that if $\int_a^b f=0$ and $f\geqslant 0$, $f=0$ on every point of continuity$^{1}$, thus no point of in $\{x:f>0\}$ can be a point of continuity. It follows $\{x:f>0\}\subset \{x:f \text{ is discontinuous at } x\}$ and by Lebesgue's criterion, $\{x:f>0\}$ has measure zero.

The converse follows since Riemann integrable implies Lebesgue integrable, for we know $f=0$ almost everywhere.

$1$. P If $x'$ is a point of continuity (we can prove such a point exists) and $f(x')>0$, we can find a closed interval $[\alpha,\beta]$ containing $x'$ where $f>0$. Then $$\int_a^b f\geqslant \int_\alpha^\beta f>0$$

Pedro
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