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To show that if $\epsilon >0$, then there exists $\delta > 0$ such that if $Q$ is any partition of $I = [a,b]$ with $||Q||< \delta$, then $L(Q;f) \geq L(f) - \epsilon$ and $U(Q;f) \leq U(f) + \epsilon$. Here $L(f) := \sup \{L(P;f), P \in \mathscr P\}$ and simillarly $U(f)$ is defined.

$\mathscr P$ is the set of all partitions of $I = [a,b]$. $f$ is a continuous function.

I am facing difficulty in doing this.

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  • $\begingroup$ What are the hypothesis on $f$? $\endgroup$ – mathcounterexamples.net Jul 13 '15 at 16:07
  • $\begingroup$ Also, what are P and $\mathscr{P}$? $\endgroup$ – gabe Jul 13 '15 at 16:58
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    $\begingroup$ It is the so called dicing lemma: see Pete Clark's Honors Calculus, p. 172 . Boundedness of $f$ is enough. $\endgroup$ – Tony Piccolo Jul 14 '15 at 3:57
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Those inequalities follow by properties of sup and inf. Namely, let $A$ be a set, if $s=\sup A$, then $s$ is the least upper bound of $A$ so for any $\epsilon\gt 0$, since the number $s-\epsilon$ is less than $s$, it is not an upper bound of $A$, so there is some $a_\epsilon\in A$ such that $a_\epsilon\gt s-\epsilon$.

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  • $\begingroup$ you missed the whole $\delta$ stuff :) $\endgroup$ – user251257 Jul 13 '15 at 19:37
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I assume you mean $$ L(Q, f) = \sum_{i=1}^{n} (x_{i} - x_{i-1}) \inf_{x_{i-1} < x < x_i} f(x) $$ for $Q = (x_0 < \dotsb < x_n)$.

Let $\epsilon > 0$. As $f$ is continuous on $[a,b]$, it is uniformly continuous. Thus, for $\eta = \frac{\epsilon}{b-a}$ there exists some $\delta > 0$ such that for every $x,x'\in[a,b]$ with $|x-x'|<\delta$ it follows $|f(x) - f(x')| < \eta$.

Let $Q = (a=x_0 < \dotsb < x_n=b)$ be a partition of $[a,b]$ with $$\|Q\| = \max_{1\le i \le n} x_i - x_{i-1} < \delta.$$ Then, we have $$\begin{align} L(f) &\le \sum_{i=1}^n (x_i - x_{i-1}) \sup_{x_{i-1} < x < x_i} f(x) \\ &\le \sum_{i=1}^n (x_i - x_{i-1}) \inf_{x_{i-1} < x < x_i} f(x) + \sum_{i=1}^n (x_i - x_{i-1}) \sup_{x_{i-1} < x,x' < x_i} |f(x) - f(x')| \\ &= L(Q,f) + \sum_{i=1}^n (x_i - x_{i-1}) \eta \\ &= L(Q,f) + (b-a) \eta \\ &= L(Q,f) + \epsilon. \\ \end{align} $$

The proof for $U(Q,f)$ is leaved as exercise ;)

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