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A bounded function $f: A \longrightarrow \mathbb{R}$ is integrable if and only if for every $\epsilon > 0$ there is a partition $P$ of $A$ such that $U(f,P) - L(f,P) < \epsilon$.

$\textbf{Proof}$

If this condition holds, it is clear that $\sup \ \{ L(f,P) \} = \inf \ \{ U(f,P) \}$ and $f$ is integrable. On the other hand, if $f$ is integrable , so that $\sup \ \{ L(f,P) \} = \inf \ \{ U(f,P) \}$, then for any $\epsilon > 0$ there are partitions $P$ and $P'$ with $U(f,P) - L(f,P') < \epsilon$. If $P''$ refines both $P$ and $P'$, it follows from lemma 3.1 that $U(f,P'') - L(f,P'') \leq U(f,P) - L(f,P') < \epsilon$.

P.S.: $A \subset \mathbb{R}^n$.

I would like to know why $\sup \ \{ L(f,P) \} = \inf \ \{ U(f,P) \}$ implies that there are partitions $P$ and $P'$ with $U(f,P) - L(f,P') < \epsilon$. Thanks in advance!

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    $\begingroup$ This just follows from the properties of $\sup, \inf$. If $x = \sup S$, then for any $\epsilon>0$ there is some $s \in S$ such that $s > x-\epsilon$. $\endgroup$ – copper.hat Jan 22 '17 at 19:55
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    $\begingroup$ Given $\epsilon > 0$, exists a partition $P'$ such that $\sup \ \{ L(f,P) \} - \frac{\epsilon}{2} < L(f,P')$, then $\sup \ \{ L(f,P) \} - L(f,P') < \frac{\epsilon}{2}$, therefore $\inf \ \{ U(f,P) \} - L(f,P') < \frac{\epsilon}{2}$ (i). Given $\epsilon > 0$, exists a partition $P''$ such that $U(f,P'') < \inf \ \{ U(f,P) \} + \frac{\epsilon}{2}$ (ii). Adding (i) with (ii), we have $\inf \ \{ U(f,P) \} + U(f,P'') - L(f,P') < \inf \ \{ U(f,P) \} + \frac{\epsilon}{2} + \frac{\epsilon}{2}$, so $U(f,P'') - L(f,P') < \epsilon$. I got it, thanks! $\endgroup$ – George Jan 22 '17 at 20:48

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