Exercise 7.6.11

Question

Finish the proof in this direction by explaining how to construct a partition $P_\epsilon$ of $[a,b]$ such that $U(f,P_\epsilon) - L(f, P_\epsilon) \leq \epsilon$.

Ulisse Mini · Accepted Answer

Let $\alpha = \frac{\epsilon}{2(b-a)}$, and $|f| < M$. Begin by defining the collection of open intervals $\mathcal{G} = \{G_1, \dots, G_N\}$ with the properties described in Exercise 7.6.9. Define $P_1$ as the partition containing all of the endpoints of $\mathcal{G}$.

Consider the collection $\mathcal{K}$ of subintervals which do not contain elements of $D^\alpha$; by Exercise 7.6.10, $f$ is uniformly $\alpha$-continuous over all sets in $\mathcal{K}$. Therefore, there exists a $\delta > 0$ so that for $x, y \in \mathcal{K}$, $\left|f(x) - f(y)\right| < \alpha$. Define $P_2$ as the partition consisting of evenly spaced points such that each subinterval has length at most $\delta$, and define $P_\epsilon$ as the common refinement of $P_1$ and $P_2$.

Let
$$U(f,P_\epsilon) - L(f, P_\epsilon) = \sum_{k \in S_1} (M_k - m_k) \Delta x_k + \sum_{k \in S_2} (M_k - m_k) \Delta x_k $$
where $S_1$ includes all of the intervals containing elements of $D^\alpha$ and $S_2$ includes all of the other subintervals. Now
$$\sum_{k \in S_1} (M_k - m_k) \Delta x_k < 2M \sum_{k \in S_1} \Delta x_k < \frac{\epsilon}{2} $$
and
$$\sum_{k \in S_2} (M_k - m_k) \Delta x_k < \frac{\epsilon}{2(b-a)}\sum_{k \in S_2} \Delta x_k \leq \frac{\epsilon}{2(b-a)} (b-a) = \frac{\epsilon}{2} $$
hence $U(f, P_\epsilon) - L(f, P_\epsilon) < \epsilon$ and $f$ is integrable.

Exercise 7.6.11

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Exercise 7.6.11

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