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That is

$f$ is integrable on $[a,b] \iff \forall\epsilon>0, \exists P$ of $[a,b]$ such that

$$U(f,P) - L(f,P) < \epsilon$$

I was thinking that a better definition would be if $U(f,P) = L(f,P)$, but I was corrected that it wouldn't work well for a curve like $f(x) = x$. The geometry just doesn't work.

On the other bound, saying that given any positive number, I can find a partition such that the difference $U(f,P) - L(f,P) < \epsilon$ is bounded doesn't feel like a strong enough condition for integrability. Isn't the goal of analysis is always to make $\epsilon$ as small as possible, and possibly $0$?

Please see my other question on A terminology to analysts for possible relevance. Thank you

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    $\begingroup$ Saying that for all $\varepsilon > 0$, there exists a partition $P$ such that $U(f,P)-L(f,P) < \varepsilon$ is saying exactly what you are looking for. No matter how small we make $\varepsilon$, we will be able to find some partition that makes the upper and lower sums differ by at most that value. $\endgroup$ – Cameron Williams Jun 26 '13 at 1:10
  • $\begingroup$ But we still an error of epsilon. I was hoping for a more error free definition. I was satisified with $\sup (L) = \inf (U)$ $\endgroup$ – Hawk Jun 26 '13 at 1:19
  • $\begingroup$ If $U(f,P) \neq L(f,P)$ for all $P$, then $U(f,P) - L(f,P) > 0$.Thus, given $\epsilon < \inf( U(f,P) - L(f,P) )$, there is no $P$ that will satisy $U(f,P)-L(f,p) < \epsilon$. So, the above two definitions mean the same thing. $\endgroup$ – AnonSubmitter85 Jun 26 '13 at 1:23
  • $\begingroup$ @CameronWilliams, want to post that as an answer? $\endgroup$ – Hawk Jun 26 '13 at 1:36
  • $\begingroup$ @sidht But you can make the error as small as you want. $\endgroup$ – Pedro Tamaroff Jun 26 '13 at 1:37
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As per my comment above: Saying that for all $\varepsilon>0$, there exists a partition P such that $U(f,P)−L(f,P)<\varepsilon$ is saying exactly what you are looking for. No matter how small we make $\varepsilon$, we will be able to find some partition that makes the upper and lower sums differ by at most that value.

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Isn't the goal of analysis is always to make $\epsilon$ as small as possible, and possibly 0?

You seem to have a misunderstanding of what $\epsilon$ stand for here. You are given $\epsilon>0$ and you want to make something smaller than this $\epsilon >0$. Maybe this can clear some of this out.


Why does $U(f,P) - L(f,P) < \epsilon$ make a good criterion for integrability?

Because

THM Let $x,y$ be arbitrary real numbers. If $x<y+\epsilon$ for each $\epsilon >0$, then $x\leq y$.

P By contradiction. Suppose $x>y$. Then $x-y>0$. Take $\epsilon=x-y$. The above gives $x<y+\epsilon=y+x-y=x$ which is impossible. It must be the case $x\leq y$.

Note that since $\sup L\leq \inf U$ is always true, the criterion gives that $\inf U\leq \sup L$ which means $\sup L=\inf U$ and $f$ is integrable.


Now, proving that for each $\epsilon >0$ there exists $P=P_\epsilon$ such that $$U(f,P)-L(f,P)<\epsilon$$ is usually easier than proving $\sup L=\inf U$ directly, in particular when the function is not given explicitly (say, if we want to prove $f$ is integrable when it is continuous) or any other cases where $f$ is in incognito.

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This is another extended comment, rather an answer.

I would just like to point out the the following two statements are totally different:

  • (1) $\forall \epsilon > 0$, $\exists$ partition $P$ such that $$U(f,P) - L(f,P ) < \epsilon$$

  • (2) $\exists$ partition $P$ such that $\forall \epsilon > 0$: $$U(f,P) - L(f,P) < \epsilon.$$

In (1), the partition $P$ depends on $\epsilon$. It might be more accurate to write $P$ as $P_\epsilon$ to remind ourselves of this fact.

In (2), there is a single partition $P$ -- independent of $\epsilon$ -- such that for all $\epsilon > 0$: $U(f,P) - L(f,P) < \epsilon$ is satisfied. In this situation, because the partition $P$ is independent of $\epsilon$, we can conclude that $U(f,P) = L(f,P)$.

Example: Contrast the following two statements. Think about how they are radically different:

  • (A) For all $x > 0$, there exists $y > 0$ such that $x < y$.

  • (B) There exists $y > 0$ such that for all $x > 0$: $x < y$.

Note that (A) is obviously true (for any $x> 0$, there's always some number that's bigger), while (B) is obviously false (there is no single number $y$ bigger than every real number).

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