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True or False. If $\int^{b}_a f > 0$, then $\exists \; [c,d] \subseteq [a,b]$ and $\delta > 0$ such that $f(x) \ge \delta$ for all $x \in [c,d]$.

1. We need to determine if true or false. How? I tried a figure from Stewart p367:

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2. How can we presage to prove the contrapositive? Is a direct proof possible?

3. I don't understand the 'indeed the negation...is a compact set'? We want $\color{red}{\neg}\exists \; █ \; [c,d] \subseteq [a,b] \wedge \delta > 0 \; █ \; \; [ \;f(x) \ge d \; \forall \, x \in [c,d] \; ]$ = $\forall \; \color{red}{\neg} \; █ \; [c,d] \subseteq [a,b] \wedge \delta > 0 \; █ \; [ \;f(x) \ge d \; \forall \, x \in [c,d] \; ]$ = $\forall \; █ \; [c,d] \supset [a,b] \vee \delta < 0 \; █ \; \color{red}{\neg} [ \;f(x) \ge d \; \forall \, x \in [c,d] \; ]$ = $\forall \; █ \; [c,d] \supset [a,b] \vee \delta < 0 \; █ \; \;f(x) < d \; \color{red}{\neg} \;\forall \, x \in [c,d] \; $ = $\forall \; █ \; [c,d] \supset [a,b] \vee \delta < 0 \; █ \; \;f(x) < d \; \; \exists x \in [c,d] \; $

What foundered? Can't have a quantifier at the end ?

Community
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    Do we have any promises about the continuity or differentiability of $f$? – Eric Towers Mar 07 '14 at 07:59
  • @EricTowers questions says nothing about this. i checked. –  Mar 07 '14 at 08:02
  • Obviously, for this statement to make sense, $f$ has to be Riemann-integrable, that is, of bounded variation. – Lutz Lehmann Mar 07 '14 at 08:03
  • The contraposition of $[c,d]⊆[a,b]$ is not $[c,d]⊃[a,b]$, there are many other ways one set can be not a subset of another set. – Lutz Lehmann Mar 07 '14 at 08:08
  • @LutzL: A Riemann-integrable function does not need to be of bounded variation. In fact there are differentiable functions not of bounded variation, and differentiable $\implies$ continuous $\implies$ Riemann integrable. I think you have antiderivatives in mind instead.... – Pete L. Clark Mar 08 '14 at 19:29
  • Yes, like $f(0)=0$, $f(x)=\sin(1/x)$ on $[0,1]$. I have to rethink my prejudices. – Lutz Lehmann Mar 08 '14 at 19:57

3 Answers3

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Negation of the conclusion

The correct formal statement of the conclusion is $$ ∃c,d,δ∀x:([c,d]⊆[a,b]∧ δ>0∧x∈[c,d]) ⟹ f(x)≥δ $$ The negation is $$ ∀c,d,δ∃x:([c,d]⊆[a,b]∧ δ>0∧x∈[c,d])∧f(x)<δ $$ where the defining conditions can again be absorbed into the quantor blocks $$ ∀[c,d]⊆[a,b],δ>0∃x∈[c,d]:f(x)<δ $$

On the proof of the statement

Now, contrary to the first lines of the proof, this does not mean that there are non-positive values for any interval $[c,d]$ since integrable functions need not be continuous everywhere. But still one can conclude that for any $δ>0$ $$ \inf_{x∈[c,d]} f(x)<δ ⟹ \inf_{x∈[c,d]} f(x)\le 0. $$ So that under the contraposition any lower Darboux sum is non-positive, and since the value of the Riemann integral is equal to the limit of the lower Darboux sums, one would conclude that $\int_a^b f(x)dx\le 0$, in contradiction to the assumption $\int_a^b f(x)dx> 0$.

Lutz Lehmann
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  • Sorry, I got a little distracted by all the quantifiers at the beginning of this answer. Now that I look back at it, it is saying exactly the same thing as mine. I upvoted it, and I encourage others to do so. – Pete L. Clark Mar 08 '14 at 20:18
  • Added headings to separate the two parts of the answer. – Lutz Lehmann Mar 08 '14 at 20:29
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As usual, the format of the question confuses me: there are many different fonts and it is not clear which of the text is yours, and where the text which is not yours comes from. That is not good in an academic context. But the question is interesting, so I'll make an an exception and answer it.

The given solution seems, in its use of the compactness of $[c,d]$, to be implicitly assuming the continuity of $f$: a continuous function on a compact set which takes arbitrarily small positive values must also take non-positive values. Since $f$ is not assumed to be continuous but only Riemann-integrable, this is not valid...but once we address this point, the solution becomes correct.

Rather, suppose $\int_a^b f > 0$, and seeking a contradiction we suppose that for all nontrivial subintervals $[c,d]$, there is no $\delta >0 $ such that $f(x) \geq \delta$ on for all $x \in [c,d]$. Then the infimum of $f$ on each subinterval is at most $0$. This implies that for any partition $\mathcal{P}$ of $[a,b]$, the lower Darboux sum $L(f,\mathcal{P})$ is at most $0$. But since $f$ is Riemann integrable, $\int_a^b f$ is the supremum of all the lower Darboux sums, so $\int_a^b f \leq 0$, contradiction.

Pete L. Clark
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I am not completely sure whether your solution is correct. It is correct if $f$ is continuous. This is the reason:

To negate the statement, it should be:

"For all intervals $[c, d]\subset [a, b]$ and $\delta >0$, there is $x\in [c, d]$ such that $f(x) < \delta$."

Now fix $[c, d]$ and let $\delta = 1/n$. Then there is a sequence $x_n \in [c, d]$ such that $f(x_n) < 1/n$. By Bolzano-Weiestrauss theorem, $\{x_n\}$ has a converges subsequence $\{x_{n_k}\}$ which converges to $x\in [c, d]$. (This is where we used the fact that $[c, d]$ is compact).

If $f$ is continuous, then we have $f(x) = \lim_{k\to \infty} f(x_{n_k}) \leq 0$ and the rest follows.

When $f$ is not continuous, one can modify the proof a little bit. Let $\int_a^b f = A >0$. Again assume that the statement is false. Let $\delta = \frac{A}{2(b-a)}$. So all for partition $P$ (as in your answer) we have $c_k \in [x_{k-1}, x_k]$ such that $f(c_k) \leq \delta$. Thus (as in your answer)

$$L(f, P) = \sum_{k=1}^n m_k (x_k - x_{k-1}) \leq \delta (b-a) = A/2\ .$$

Thus $\int_a^b f = \sup_P L(f, P) \leq A/2 <A$, which is a contradiction.

I do not think there is a constructive way to prove this, as there does not seem to be a way to find that interval $[c, d]$.

  • If f is continuous, then let $δ=\max f(x)/2$, let $x_=\arg\max f(x)$. Then by continuity there is a non-trivial interval $[c,d]$, $d>c$, around $x_$ with $f(x)\ge δ$ for all $x\in [c,d]$. – Lutz Lehmann Mar 07 '14 at 08:12
  • @LutzL: You are correct that if $f$ is continuous then the statement can be proved quite easily. I just want to explain why the compactness is mentioned in the solution. Actually I am still unsure whether the proof mentioned in the OP solution is valid. –  Mar 07 '14 at 08:16
  • In more abstract terms, one could say that if $\int_a^b f(x)dx>0$, then the set ${x:f(x)>0}$ has positive (Jordan) measure and thus must contain inner points, and consequently open intervals, and making them smaller, also closed intervals. – Lutz Lehmann Mar 07 '14 at 08:24
  • @LutzL: Sorry that I am not familiar with Jordan measure, but note that we'd better not want to go away from the Riemann world, as one can always change the definition of $f$ on a set of measure $0$ such that the statement cannot be true. –  Mar 07 '14 at 08:29
  • Jordan measure is actually the measure for Riemann integrals, what you mean is Borel or Lebesgue measure. And I think one has to go your way also for continuous functions, since one would have to explain why my $δ$ is positive. The proof is correct, for R-integrable functions the limit over the lower Darboux sums is the value of the integral is the limit over all Riemann sums is the limit over the upper Darboux sums. And by contraposition, any lower Darboux sum is non-positive. – Lutz Lehmann Mar 07 '14 at 08:33
  • @LutzL: Thanks for your comment. Actually in the solution I am not quite sure why the negation would be of the form described (First three lines of the answer) –  Mar 07 '14 at 08:47
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    For real functions in one variable, Riemann-integrable is the same as possessing left and right limits in every point. This would include examples like $f(x)=|x|$ for $x\ne 0$ and $f(0)=1$, so on $[-1,1]$ there is some $x_n$ with $f(x)<1/n$ for every $n$, but $f(x)>0$ everywhere. So indeed, how would that fit with those first three lines? But still, what is important is that $0=\inf{ f(x): x\in [-1,1]}$. – Lutz Lehmann Mar 07 '14 at 08:55
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    @LutzL: Riemann integrable is not the same as possessing left and right limits at every point. That condition is called "regulated", see e.g.http://en.wikipedia.org/wiki/Regulated_function. For instance the function $f(0) = 0$, $f(x) = \sin(\frac{1}{x})$ for $0 < x \leq 1$ is bounded on $[0,1]$ and discontinuous only at $0$, so Riemann integrable, but it is not regulated. Moreover regulated functions have only countably many discontinuities, whereas Riemann integrable functions can have uncountably many, so long as the set has measure zero (and the function is bounded). – Pete L. Clark Mar 08 '14 at 19:45