Let $f$ be a real-valued function on $[a,b]$. Assume $f$ is Riemann integrable with strictly positive Riemann integral over $[a,b]$ then $f$ is strictly positive on some nonempty open interval.

What if Lebesgue integrable instead of Riemann?

*What I'm thinking is the following for Riemann.

Assume otherwise, then for all open interval, say $(a',b')$,$\int_{a'}^{b'}f(x) dx\leq 0$. Since $f$ is Riemann-integrable with strictly positive Riemann integral over $[a,b]$, $\int_a^b f(x)dx< 0$. Let $a'\rightarrow a$, and $b'\rightarrow b $, then $\forall\epsilon>0$, $\exists\delta>0$ such that if $|a-a'|<\delta$ and $|b-b'|<\delta$, then $|\int_a^b f(x)dx-\int_{a'}^{b'}f(x)dx|<\epsilon$. By letting $\epsilon\rightarrow 0$, we'll have $\int_a^bf(x)dx\leq 0$, contradiction.

Is this correct? and what about Lebesgue?

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    $\begingroup$ Why is $\int_{a'}^{b'} f(x)dx \le 0$? This does not follow $\endgroup$ – mathworker21 Aug 1 '17 at 0:52
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    $\begingroup$ shouldn't it be: assume otherwise, then for every open interval $(a',b')$, $f$ is not strictly positve on that interval? (that doesn't imply that the integral over that interval is negative) $\endgroup$ – Jens Renders Aug 1 '17 at 0:55
  • $\begingroup$ Typically, for lebesgue integration you should expect things to happen "almost surely". $\endgroup$ – N. S. Aug 1 '17 at 1:03
  • $\begingroup$ @N.S. Was the proof ok for Riemann? Thank you! $\endgroup$ – 2ndaccount Aug 1 '17 at 1:15
  • $\begingroup$ At the end of the second sentence of the proof, should $\int_a^b f(x)dx< 0$ be $\int_a^b f(x)dx > 0$? $\endgroup$ – saulspatz Aug 1 '17 at 1:51

The theorem is false in the Lebesgue setting. Consider $$f(x) = \begin{cases} 0 & x \in \mathbb{Q} \\ 1 & \textrm{otherwise} \end{cases} .$$ Your very first sentence of your "proof" is wrong.

EDIT: I just realized this was meant to be a proof for Riemann-integrable functions, not Lebesgue.

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    $\begingroup$ The OP's proof was for Riemann integrals, and it seems to be fine. Then the OP's other question was whether the same theorem is true for Lebesgue integrals. $\endgroup$ – zipirovich Aug 1 '17 at 1:04

The proof for Riemann could be easier.

Since $\int_a^bf(x)dx>0$, there is a Partition $P$ such that $L(P,f)>0$, where $L(P,f)$ means Darboux lower sum. So there must be a interval, where $\inf f>0$


For Lebesgue integrals, this is not even true up to modification on a set of measure zero. Take for instance the indicator function of a "fat" Cantor set (cf. page 39 of Folland's 'Real Analysis').

  • $\begingroup$ Wikipedia describes such a set. It's indicator function is a good counter-example. $\endgroup$ – Gribouillis Aug 1 '17 at 9:17

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