Let $f : Q\to \mathbb{R}$ be bounded. Then the statement that $f$ is integrable over $Q$, with $\int_{Q}f = A$, is equivalent to the statement

Question

Theorem. Let $f : Q\to \mathbb{R}$ be bounded. Then the statement that $f$ is integrable over $Q$, with $\int_{Q}f = A$, is equivalent to the statement that given $\epsilon>0$, there is a $\delta>0$ such that if $P$ is any partition of mesh less than $\delta$, and if, for each subrectangle $R$ determined by $P$, $x_R$ is a point of $R$, then $|\sum_{R}f(x_R)v(R) - A|<\epsilon$.

The exercise of which I am doing one is the following: 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 that still has no answer and I would like to know how to demonstrate that too.

Suppose that $f$ is integrable with $\int_{Q}f$, then by the exercise, given $\epsilon>0$ there exists a $\delta>0$ such that if $P$ is a partition with a norm smaller than $\delta$, we have that $U(f,P)-L(f,P)<\epsilon$. Let $x_R\in R$ then we have the following inequalities $\sum_{R}f(x_R)v(R)\leq U(f,P)$ and $A\geq L(f,P)$ and so $\sum_{R}f(x_R)v(R)-A\leq U(f,P)-L(f,P)<\epsilon$, but I do not know how to prove this so I can keep the absolute value. Could anyone help me, please?

For the other direction, suppose the second part of the theorem, then we will have that $U(f,P)-L(f,P)\leq |U(f,P)-L(f,P)|\leq |U(f,P)-A|+|L(f,P)-A|<2\epsilon$ and thus $f$ is integrable.

Answer 1

You have correctly shown that for any partition $P$ with $\|P\| < \delta$ and for any Riemann sum $\sum_R f(x_R) v(R)$ correspnding to $P$, it is true that

$$\tag{1}\sum_R f(x_R) v(R)-A < \epsilon,$$

where $A = \int_Q f.$

Similarly, we have $A \leqslant U(f,P)$ and $L(f,P) \leqslant \sum_R f(x_R) v(R)$ and it follows that

$$A - \sum_R f(x_R) v(R) \leqslant U(f,P) - L(f,P) < \epsilon.$$

This implies $-\epsilon < \sum_R f(x_R) v(R) - A$ and along with (1) we have

$$\tag{2}-\epsilon < \sum_R f(x_R) v(R) - A< \epsilon.$$

Now (2) allows us to "keep the absolute value":

$$\left|\sum_R f(x_R) v(R) - A \right| < \epsilon$$

Answer 2

I am reading James R. Munkres "Analysis on Manifolds" now.
I solved exercise 7(p.90).

$(\Rightarrow)$

Let $\epsilon$ be an arbitrary positive real number.
By exercise 6(p.90), there exists $\delta > 0$ such that $U(f, P) - L(f, P) < \epsilon$ for any partition $P$ whose mesh is less than $\delta$.
Let $P$ be an arbitrary partition whose mesh is less than $\delta$.
Because $A = \inf_P \{U(f, P)\} = \sup_P \{L(f, P)\}$,
$$ L(f, P) \leq A \leq U(f, P) $$ and $$ L(f, P) \leq \sum_R f(x_R) v(R) \leq U(f, P). $$ So, $$ \left|\sum_R f(x_R) v(R) - A \right| \leq U(f, P) - L(f, P) < \epsilon. $$

$(\Leftarrow)$
Let $\epsilon$ be a positive real number.
Let $P$ be an arbitrary partition such that
$$ A - \epsilon < \sum_R f(x_R) v(R) < A + \epsilon, $$ where $x_R$ is an arbitrary point in $R$.
Then $U(f, P) \leq A + \epsilon$.
The reason is the following:
Assume $A + \epsilon < U(f, P)$.
Let $\epsilon^{'} = \frac{\sum_R M_R(f) v(R) - A - \epsilon}{v(Q)}$.
For each $R$, there exists $x_R$ such that
$$ M_R(f) - \epsilon^{'} < f(x_R) \leq M_R(f). $$ Then, $$ \sum_R (M_R(f) - \epsilon^{'}) v(R) < \sum_R f(x_R) v(R) \\ \sum_R M_R(f) v(R) - \epsilon^{'} \sum_R v(R) < \sum_R f(x_R) v(R) \\ \sum_R M_R(f) v(R) - \epsilon^{'} v(Q) < \sum_R f(x_R) v(R) \\ A + \epsilon < \sum_R f(x_R) v(R). \\ $$ This is a contradiction.
Similarly,
$A - \epsilon \leq L(f, P)$.
So,
$$ A - \epsilon \leq L(f, P) \leq \sup_P \{L(f, P)\} \leq \inf_P \{U(f, P)\} \leq U(f, P) \leq A + \epsilon \\ 0 \leq \inf_P \{U(f, P)\} - \sup_P \{L(f, P)\} \leq 2 \epsilon \\ \therefore \inf_P \{U(f, P)\} = \sup_P \{L(f, P)\} $$ and $$ -\epsilon \leq \sup_P \{L(f, P)\} - A \leq \epsilon \\ \left| \sup_P \{L(f, P)\} - A \right| \leq \epsilon \\ \therefore \sup_P \{L(f, P)\} = A $$