Exercise 8.1.1

Question

\begin{enumerate}[label=(\alph*)] 
\item Explain why both the Riemann sum $R(f,P)$ and $\int^b_a f$ fall between $L(f,P)$ and $U(f,P)$.
\item Explain why $U(f,P') - L(f,P') < \epsilon / 3$.
 \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
\item $L(f,P) \leq R(f,P) \leq U(f,P)$ is clear from their definitions, as noted earlier in the section's discussion. The definition of $\int^b_a f$ as the supremum of $L(f,P)$ over all partitions $P$ shows $\int^b_a f \geq L(f,P)$, and similar reasoning gives $\int^b_a f \leq U(f,P)$.
\item $P$ is a refinement of $P_\epsilon$, so from Lemma 7.2.3,
$$U(f, P') - L(f,P') = U(f,P_\epsilon) - L(f, P_\epsilon) < \frac{\epsilon}{3}$$
 \end{enumerate}

Exercise 8.1.1

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Exercise 8.1.1

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