Riemann integrable vs Lebesgue integrable

Question

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?

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) — Jens Renders, Aug 01 '17 at 00:55
Typically, for lebesgue integration you should expect things to happen "almost surely". — N. S., Aug 01 '17 at 01:03
At the end of the second sentence of the proof, should $\int_a^b f(x)dx< 0$ be $\int_a^b f(x)dx > 0$? — saulspatz, Aug 01 '17 at 01:51
Your first line of proof is wrong. The exact counter statement is that for every open interval $I$ there is a point $x\in I$ such that that $f(x) \leq 0$. You on the other hand think that $f\leq 0$ on whole of $I$. — Paramanand Singh, Aug 01 '17 at 01:52
@ParamanandSingh Then how should I prove the statement for Riemann? Thank you! — secondaccount, Aug 01 '17 at 01:56
An answer here uses Darboux sums. And in comment to that answer I have linked another proof. — Paramanand Singh, Aug 01 '17 at 02:04

score 10 · Answer 1 · edited Aug 01 '17 at 03:27

10

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 $\inf f>0$

edited Aug 01 '17 at 03:27

AJY

8,729

answered Aug 01 '17 at 01:08

Selene

1,093

This is much better than Apostol's proof (https://math.stackexchange.com/a/2350319/72031) +1 – Paramanand Singh Aug 01 '17 at 02:03

AJY · Accepted Answer · 2017-08-01T01:05:15.270

6

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.

edited Aug 01 '17 at 01:05

answered Aug 01 '17 at 01:00

AJY

8,729

1

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. – zipirovich Aug 01 '17 at 01:04

score 4 · Answer 3 · answered Aug 01 '17 at 01:22

4

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').

answered Aug 01 '17 at 01:22

shalop

13,703

Wikipedia describes such a set. It's indicator function is a good counter-example. – Gribouillis Aug 01 '17 at 09:17

Riemann integrable vs Lebesgue integrable

3 Answers3