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