# Let $f : Q\to \mathbb{R}$ be bounded. Then $f$ is integrable over $Q$ if and only if given $\epsilon> 0$, there is a $\delta> 0$ such that

Prove the following:

Theorem. Let $f : Q\to \mathbb{R}$ be bounded. Then $f$ is integrable over $Q$ if and only if given $\epsilon> 0$, there is a $\delta> 0$ such that $U(f, P) -L(f, P) <\epsilon$ for every partition $P$ of mesh less than $\delta$.

Proof. (a) Verify the "if" part of the theorem.

(b) Suppose $|f(x)| < M$ for $x\in Q$. Let $P$ be a partition of $Q$. Show that if $P"$ is obtained by adjoining a single point to the partition of one of the component intervals of $Q$, then

$$0\leq L(f, P") - L(f, P)\leq 2M(\text{mesh} P) (\text{width} Q)^{n-1}$$

Derive a similar result for upper sums.

(c) Prove the "only if" part of tbe theorem: Suppose $f$ is integrable over $Q$. Given $\epsilon> 0$, choose a partition $P'$ such that $U(f, P')- ­ L(f, P') < \epsilon/2$. Let $N$ be the number of partition points in $P'$; then let

$$\delta= \epsilon/8M N (\text{width} Q)^{n-1}$$

Show that if $P$ has mesh less than $\delta$, then $U(f, P) - L(f, P) <\epsilon$.

[Hint: The common refinement of $P$ and $P'$ is obtained by adjoining at most $N$ points to $P$.]

(a) if $\epsilon>0$ then there is a $\delta>0$ such that $U(f,P)-L(f,P)<\epsilon$ for any partition of norm smaller than $\delta$, if $P$ a partition of norm smaller than $\delta$, then $U(f,P)-L(f,P)<\epsilon$ and thus $f$ is integrable.

(b) I know that if $P''$ is a partition finer than $P$ then we have to $L(f,P)\leq L(f,P'')$, with which $0\leq L(f,P'')-L(f,P)$. I am entangled in testing the other inequality, I know that $L(f,P'')-L(f,P)=\sum_{R\subset P''}m_R(f)v(R)-\sum_{R\subset P}m_R(f)v(R)$, but I do not know how to limit that, could someone help me please?

(c) I do not know how to prove this, could someone give me a help or do a test? Thank you very much.

• I am reading James R. Munkres "Analysis on Manifolds" now. And I was able to solve exercise 6(b)(p.90) by reading the proof of Lemma 10.1(p.83) again. – tchappy ha Sep 5 '18 at 12:20

James R. Munkres "Analysis on Manifolds"
p.90 Exercise 6 (b)

Let $Q$ be the rectangle

$$Q = [a_1, b_1] \times \cdots \times [a_n, b_n]$$

Let $P^{''}$ be the partition obtained by adjoining a single additional point to the partition of one of the component intervals of $Q$.

Suppose, to be definite, that $P$ is the partition $(P_1, \cdots, P_n)$ and that $P^{''}$ is obtained by adjoining the point $q$ to the partition $P_1$.

Further, suppose that $P_1$ consists of the points

$$a_1 = t_0 < t_1 < \cdots < t_k = b_1$$

and that $q$ lies interior to the subinterval $[t_{i-1}, t_i]$.

We first compare the lower sums $L(f, P)$ and $L(f, P^{''})$. Most of the subrectangles determined by $P$ are also subrectangles determined by $P^{''}$. An exception occurs for a subrectangle determined by $P$ of the form

$$R_S = [t_{i-1}, t_i] \times S$$

(where $S$ is one of the subrectangles of $[a_2, b_2] \times \cdots \times [a_n, b_n]$ determined by $(P_2, \cdots, P_n)$). The term involving the subrectangle $R_S$ disappears from the lower sum and is replaced by the terms involving the two subrectangles

$$R_S^{'} = [t_{i-1}, q] \times S \text{ and } R_S^{''} = [q, t_i] \times S \text{,}$$

which are determined by $P^{''}$.

$$0 \leq L(f, P^{''}) - L(f, P) \\ = \sum_{S} [m_{R_S^{'}}(f) v(R_S^{'}) + m_{R_S^{''}}(f) v(R_S^{''})] - \sum_{S} m_{R_S}(f) v(R_S)\\ = \sum_{S} [m_{R_S^{'}}(f) v(R_S^{'}) + m_{R_S^{''}}(f) v(R_S^{''})] - \sum_{S} m_{R_S}(f) (v(R_S^{'}) + v(R_S^{''})) \\ = \sum_{S} (m_{R_S^{'}}(f) - m_{R_S}(f)) v(R_S^{'}) + \sum_{S} (m_{R_S^{''}}(f) - m_{R_S}(f)) v(R_S^{''}) \\ \leq \sum_{S} 2 M v(R_S^{'}) + \sum_{S} 2 M v(R_S^{''}) \\ = 2 M \sum_{S} [v(R_S^{'}) + v(R_S^{''})] \\ = 2 M \sum_{S} v(R_S) \\ = 2 M (t_i - t_{i-1}) (b_2 - a_2) \cdots (b_n - a_n) \\ \leq 2 M (\text{mesh } P)(\text{width } Q)^{n-1} \text{.}$$

$$0 \leq U(f, P) - U(f, P^{''}) \\ = \sum_{S} M_{R_S}(f) v(R_S) - \sum_{S} [M_{R_S^{'}}(f) v(R_S^{'}) + M_{R_S^{''}}(f) v(R_S^{''})] \\ = \sum_{S} M_{R_S}(f) (v(R_S^{'}) + v(R_S^{''})) - \sum_{S} [M_{R_S^{'}}(f) v(R_S^{'}) + M_{R_S^{''}}(f) v(R_S^{''})] \\ = \sum_{S} (M_{R_S}(f) - M_{R_S^{'}}(f)) v(R_S^{'}) + \sum_{S} (M_{R_S}(f) - M_{R_S^{''}}(f)) v(R_S^{''}) \\ \leq \sum_{S} 2 M v(R_S^{'}) + \sum_{S} 2 M v(R_S^{''}) \\ = 2 M \sum_{S} [v(R_S^{'}) + v(R_S^{''})] \\ = 2 M \sum_{S} v(R_S) \\ = 2 M (t_i - t_{i-1}) (b_2 - a_2) \cdots (b_n - a_n) \\ \leq 2 M (\text{mesh } P)(\text{width } Q)^{n-1} \text{.}$$

James R. Munkres "Analysis on Manifolds"
p.90 Exercise 6 (c)
The following answer is from "Analysis on Manifolds Solution of Exercise Problems" written by Yan Zeng.

Let $\epsilon$ be an arbitrary positive real number.
Let $P^{'}$ be a partition such that $U(f, P^{'}) - L(f, P^{'}) < \frac{\epsilon}{2}$.
Let $N$ be the number of partition points in $P^{'}$.
Let $\delta := \frac{\epsilon}{8 M N (\text{width } Q)^{n-1}}$.
Let $P$ be an arbitrary partition which has mesh less than $\delta$.
Let $P^{''}$ be the common refinement of $P$ and $P^{'}$.

Then, $P^{''}$ is obtained by adjoining at most $N$ points to $P$.

Using (b),

$$0 \leq L(f, P^{''}) - L(f, P) \leq N 2 M (\text{mesh } P) (\text{width } Q)^{n-1} < N 2 M \frac{\epsilon}{8 M N (\text{width } Q)^{n-1}} (\text{width } Q)^{n-1} = \frac{\epsilon}{4}.$$

Similarly,

$$0 \leq U(f, P) - U(f, P^{''}) < \frac{\epsilon}{4}.$$

$$U(f, P) - L(f, P) = [U(f, P) - U(f, P^{''})] + [L(f, P^{''}) - L(f, P)] + [U(f, P^{''}) - L(f, P^{''})] < \frac{\epsilon}{4} + \frac{\epsilon}{4} + [U(f, P^{'}) - L(f, P^{'})] < \frac{\epsilon}{4} + \frac{\epsilon}{4} + \frac{\epsilon}{2} = \epsilon.$$