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The theorem states

If $f$ is bounded on $[a,b]$, then $f$ is integrable on $[a,b]$ $\iff$ for all $\epsilon > 0$ there is a partition $P$ of $[a,b]$ such that

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

The proof is on page 220. I believe I have the 2nd ed of the book.

One of the steps really confused me in which he writes

"$\inf \{ U(f,P')\} - \sup\{ L(f,P')\} < \epsilon$, since this is true for all $\epsilon > 0$, it follows that $\inf \{ U(f,P')\} = \sup\{ L(f,P')\}$

I do not understand how this follows, how did this become an equality?

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  • $\begingroup$ So he is saying that $\lim_{x \to a} f(x) - L = 0 \implies f(x) = L$? $\endgroup$ – Hawk Mar 10 '13 at 20:16
  • $\begingroup$ Actually why is the LHS bounded below by 0? $\endgroup$ – Hawk Mar 10 '13 at 20:17
  • $\begingroup$ No that's exactly what is in the book. Though the inf of a set is a number and so is the sup. So we are really subtracting numbers here $\endgroup$ – Hawk Mar 10 '13 at 20:24
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This is a cute strategy in analysis: to prove that $x = y$, you instead prove that $|x - y| < \epsilon$ for all $\epsilon > 0$. Because if $x \neq y$, then taking $\epsilon = |x - y|$ would falsify the inequality. In your case, $x = \inf U(f,P')$ and $y = \sup L(f,P')$.

In your question, the absolute values are omitted because $x > y$ already; this is because we always have $U(f,P') \geq L(f,P')$ for the same partition, which is the intuition behind the use of "upper" and "lower" in the names of these quantities.

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    $\begingroup$ But he didn't use absolute values at all $\endgroup$ – Hawk Mar 10 '13 at 20:32
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    $\begingroup$ Absolute values aren't needed because the difference in question is non-negative. All the numbers $U(f,P)$ (as $P$ varies) are $\geq$ all the numbers $L(f,P)$, so the infimum of the former is at least as large as the supremum of the latter. $\endgroup$ – Andreas Blass Mar 10 '13 at 20:34
  • $\begingroup$ Can expand on the $\epsilon = |x - y| $ part a bit? I am still have trouble seeing the conclusion $\endgroup$ – Hawk Mar 10 '13 at 21:06
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    $\begingroup$ @sizz: Suppose you know that $|x-y| \lt \epsilon$ for all $\epsilon > 0$. If $x \ne y$, you can take $\epsilon = |x-y|$ (because $x-y$ is not $0$), and you get $|x-y| < |x-y|$, a contradiction which came from assuming $x \ne y$. $\endgroup$ – Javier Mar 10 '13 at 21:45
  • $\begingroup$ Oh I see, he omitted a lot of steps here... $\endgroup$ – Hawk Mar 10 '13 at 22:21
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This is called a Darboux integral and to show that $\inf \{ U(f,P')\} \geq \sup\{ L(f,P')\}$ you first have to show that for a refinement $P'$ of a partition $P$ you always have $$U(f,P) \geq U(f,P') \geq L(f,P') \geq L(f,P).$$ Then to show that all the upper integrals are bigger than all the lower integrals you can use the fact that for any two partitions $P_1$ and $P_2$ there is always a partition $P'$ that is a refinement of both of them, $P' = P_1 \cup P_2$ for example.

Now assume we have $0 \leq \inf \{ U(f,P')\} - \sup\{ L(f,P')\} < \epsilon$ for every $\epsilon > 0$ as assumed in the question. That doesn't leave room for the difference to be anything other than zero since $\epsilon$ can be chosen arbitrarily small.

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  • $\begingroup$ How did you turn a strict inequality into an ordinary inequality? $\endgroup$ – Hawk Mar 10 '13 at 21:47
  • $\begingroup$ @sizz Typo. Thanks for pointing it out, I've fixed it. $\endgroup$ – Gibarian Mar 10 '13 at 21:52

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