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I've been struggling with this for a while, and I have a couple of leads that kind of got me nowhere:

At first I thought that if $f$ is continuous somewhere then the integral will be $>0$. So, if the integral was $0$ then that would mean it would need to be nowhere continuous. That seemed unlikely to me, but I couldn't prove the existence of a point at which it is continuous.

For the integral to be $0$ it would necessitate that for any sub interval of $[a,b]$ the function's infimum would have to be $0$. Also seems weird for $f>0$. Again, got me nowhere.

I should mention that I'm aware that it's possible to prove this by defining "Measure" and all that, but I don't want to go there. I'm wondering if there are more elementary tools to show the above.

Thank you!

Adar Hefer
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    Do you mean Riemann integrable? – Paul Gustafson Apr 10 '13 at 19:09
  • I do mean Riemann integrable, thanks. I'll add it to the post. – Adar Hefer Apr 10 '13 at 19:10
  • If it's Riemann integrable, then it's not nowhere continuous – Cocopuffs Apr 10 '13 at 19:11
  • @Cocopuffs What? =O – Pedro Apr 10 '13 at 19:12
  • I was told that by my teaching assistant, but that relied on a theorem involving the definition of Measure 0. – Adar Hefer Apr 10 '13 at 19:12
  • @Cocopuffs, In fact, it's continuous a.e., but he doesn't want to use that. – Paul Gustafson Apr 10 '13 at 19:12
  • @kahen I searched for it before I posted, didn't find anything - but then again I wasn't very patient and didn't look through many threads.. – Adar Hefer Apr 10 '13 at 19:15
  • I think really the most natural and clear proof is to use the fact that Riemann integrable functions are continuous except on a set of exterior measure 0. It does not take long to prove this theorem. – Paul Gustafson Apr 10 '13 at 19:20
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    OP: The best way to search MSE in my experience is to use google's site: functionality. Try clicking this link. – kahen Apr 10 '13 at 19:21
  • Thanks for the tip. :) I really did use the site's functionality. Anyway I checked out the link you provided with a similar question, but that one involved $f$ which is continuous, which isn't the case I'm considering. @user66345 It's just that we were given this question as a starred question in our homework assignment, and I'm quite convinced we should be able to solve it without resigning to material not-yet-taught in class. – Adar Hefer Apr 10 '13 at 19:24
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    If this ends up closed as a duplicate, I encourage you to ask again, put a link to the other question in your new question, and remember to write that you specifically are interested in an answer that doesn't use measure theory. – kahen Apr 10 '13 at 19:41
  • Could you please clarify and rewrite, for example the following passage isn't clear to me:"At first I thought that if f is continuous somewhere then the integral will be $>0$. So, if the integral was $0$ then that would mean it would need to be nowhere continuous. That seemed unlikely to me, but I couldn't prove the existence of a point at which it is continuous." Take $f(x)=0$ then $\int_a^b f(x)=0$ but this is smooth/continuous everywhere. – Squirtle Apr 10 '13 at 22:56
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    @dustanalysis That particular passage is quite clear, since it is already in the context that $f>0$. If $f(x_0)>0$ and $f$ is continuous at $x_0$ then $f(x)$ can be bounded away from $0$ on a neighbourhood of $x_0$. – Erick Wong Apr 11 '13 at 01:51

2 Answers2

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Sketch: For a bounded function $f$, we define the oscillation of $f$ at $a$ by $$\mathrm{osc}f(a) = \limsup_{x\to a} f(x) - \liminf_{x\to a} f(x). $$ Then it is easy to show that $f$ is continuous at $a$ if and only if $\mathrm{osc}f(a) = 0$.

Now let us given $\epsilon > 0$ and $\delta > 0$. Then we can find a partition $P$ such that $U(f,P)-L(f,P) < \epsilon$. Thus the set $S = S(\epsilon, \delta)$ denotes the collections of subintervals $I$, formed by the points of $P$, where $\sup_{I} f - \inf_{I} f > \delta$ satisfies

$$ \epsilon > U(f,P)-L(f,P) > \sum_{I\in S} \delta |I| $$

Now, choose $(\epsilon_n, \delta_n)$ such that $\delta_n \downarrow 0$ and

$$ l := \sum_{n=1}^{\infty} \frac{\epsilon_n}{\delta_n} < b-a. $$

Choose $r >1$ such that $rl < b-a$. For each closed interval $I \in S(\epsilon_n, \delta_n)$, we choose an open interval $J$ containing $I$ such that $|J| = r|I|$. Finally, let $\mathcal{U}$ be the family formed by collecting all such open intervals $J$. If $f$ has no point of continuity, then each $x \in [a, b]$ is a point where $f$ has positive oscillation and hence lies in some $J\in \mathcal{U}$. Thus $\mathcal{U}$ is an open cover of $[a,b]$, and hence it has a finite subcover. But the sum of the length of the open intervals in that subcover cannot exceed $rl < b-a$, a contradiction! Therefore $f$ must have a point of continuity.

Of course, this proof also hides the idea of measure, though indirectly and elusively.

Sangchul Lee
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  • Thanks! I see this question was re-opened somehow. I actually got an answer on the other thread that I opened because this one was wrongfully closed. Thanks for the proof, it is a little different than the one I saw on the other thread and very interesting. – Adar Hefer Apr 15 '13 at 12:43
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What does it mean for it to be Riemann integrable? Why $should$ it be positive? The Riemann sum is defined as $$\sum_{i=0}^{n-1} f(t_i)(x_{i+1}-x_i) : t_i \in (x_{i+1},x_i)$$ over a partition $a=x_0 <x_1 < x_2 < \cdots < x_n = b$

and what is the Riemann integral? In essence, we let $n\to \infty$ (our partition becomes very fine). So essentially your question is logically equivalent to asking why the above sum is always positive, well that's because $f(t_i)>0$, so naturally the sum is too... and the integral version just gives us the best solution (but it will always be positive for the same reason).

Squirtle
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  • Oh, ... I just realized that your question has little to do with the title.... so you need to change the title. – Squirtle Apr 10 '13 at 20:32
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    well ,the upper sum is always positive but we are interested in the limit, as you know the limit of positives may be zero.. – Halil Duru Apr 11 '13 at 00:09
  • Can't we also say that the Riemann integral is bounded BELOW by the lower Riemann sum (which is positive for the same reason). ... notice, I never clearly stated where $t_i$ lies in $(x_{i+1},x_i)$. We may choose it in the worse possible way (lowest lower bound for the sum) and it's still positive. I think my logic is right, no? – Squirtle Apr 11 '13 at 00:43
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    @dustanalysis By "lower Riemann sum" do you mean "lower Darboux sum"? There may not be a minimum value to give a lowest possible bound. How do you know that the infimum of $f$ on $(x_{i-1},x_i)$ is positive? – Erick Wong Apr 11 '13 at 01:48
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    $m_i$=inf {$x\in[x_i-1,x_i]$ :f(x)} -- so this corresponds to OP's infimum of every subinterval must be zero.We must somehow contradict this with Riemann integrability.. – Halil Duru Apr 11 '13 at 05:20
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    If $L(f, P)$ is a lower sum then $$L(f,P) \le \int_a^b f dx.$$ Each lower sum is clearly positive. –  Apr 12 '13 at 08:21
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    @PeteMarkou Have you actually tried to prove that each lower sum is positive before calling it clear? This is precisely what OP is asking for. For instance, if $f(x)=1$ for irrational $x$ and $f(x)=1/n$ for $x=m/n$ (lowest terms), then $f>0$ but no lower sum is positive (of course $f$ is not Riemann-integrable). – Erick Wong Apr 12 '13 at 12:37