Riemann integral property of partitions Hello friends could you help me with the following statement please, it is still not clear to me how to use partitions and what is necessary to demonstrate the property of Riemann integrals:
Let be $P_{n}$ a partition of the rectangle $R=[0,1]\times[0,1]$ into $(2^{n})^{2}$ rectangles given by the lines $x_{i}=\frac{i}{2^{n}},y=\frac{j}{2^{n}}$ with $i,j=0,1,...,2^{n}$. Let $f:R \rightarrow \mathbb{R}$ be a bounded function. Show that $f$ is integrable if and only if $\lim L(f,P_{n})=\lim U(f,P_{n})$ as $n\rightarrow \infty$ why do these limits always exist?
As I mention above, I have tried to do it with higher and lower sums, it is clear to me that I must demonstrate and even drawing I see it clearly, but I do not know how to write it, I would greatly appreciate it. Regards.
 A: The reverse implication is that existence of the integral implies $\lim_{n \to \infty}L(f,P_n) = \lim_{n \to \infty}U(f,P_n)$ for this specific sequence of (dyadic) partitions.  The proof must rely on the fact that the partition norm $\|P_n\| \to 0$ as $n \to \infty$, since that implication is not true for partition sequences in general.
The most common definition of the Riemann integral is that there exists $I$ and, for any $\epsilon > 0$, there exists a partition $P_\epsilon$ such that for any refinement $P \supset P_\epsilon$ we have $|S(f,P) - I| < \epsilon$ for any choice of tags in the Riemann sum $S(f,P)$. Under this definition, we have the Riemann criterion that $f$ is Riemann integrable if for any $\epsilon > 0$ there exists partition $P_\epsilon$ such that $U(f,P_\epsilon) - L(f,P_\epsilon) < \epsilon$. It easily follows that for any refinement $P \supset P_\epsilon$ we have $U(f,P) - L(f,P) < \epsilon$ as well.
The Riemann criterion alone will not facilitate proof here.  It is true that the sequences of Darboux sums $(L(f,P_n))$ and $(U(f,P_n))$ are nondecreasing and nonincreasing, respectively. These sequences are bounded, since for all $n \in \mathbb{N}$
$$L(f,P_n) \leqslant \int_a^b f(x) \, dx \leqslant U(f,P_n)$$
Therefore, as these are bounded monotone sequences, the limits exist with
$$\lim_{n \to \infty}L(f,P_n) \leqslant \int_a^b f(x) \, dx \leqslant \lim_{n \to \infty}U(f,P_n)$$
We also know that there exists some partition $P_\epsilon$ such that for any refinement $P \supset P_\epsilon$ we have $U(f,P) - L(f,P) < \epsilon$.  If it were known that there existed a partition of the form $P_n$ that refined $P_\epsilon$, then the proof would be finished, as we would have $\lim_{n \to \infty}U(f,P_n) - \lim_{n \to \infty}L(f,P_n) < \epsilon$ for any $\epsilon > 0$.
However, this is not guaranteed as the partition points in $P_n$ are all dyadic rationals and $P_\epsilon$ might have non-dyadic rational points.
Fortunately, there is an equivalent criterion whereby the integral exists if for any $\epsilon > 0$ there exists $\delta > 0$ such that $U(f,P) - L(f,P) < \epsilon$ when $\|P\| < \delta$.  This is somewhat difficult  to prove  with the common definition of the Riemann integral as the starting point, particularly for the multidimensional version.  A proof along these lines is given here.
Knowing this alternative criterion it is easy to finish the proof, because $\|P_n\| = 2^{-n} \to 0$ as $n \to \infty$.
A: Existence of these limits actually follows directly from $f$ being bounded on $R$.
Let's do from the $(\implies)$ way first:
Suppose $f$ is integrable, then
$$\underline I_R(f) = \overline I_R(f)$$ in other words
$$\sup_P L(f,P_n) = \inf_P U(f,P_n)$$
We know that if $P'$ is a refinement of $P$, then $L(f,P') \geq L(f,P)$ and $U(f,P') \leq L(f,P)$. Since $P_{n+1}$ is a refinement of $P_n$ for any $n\in\mathbb{R}$ while $n\to\infty$ we shall have that
$$L(f,P_n) = \underline I_R(f) \quad \text{and}\quad U(f,P_n) = \overline I_R(f)$$ which is desired to be shown.
$(\impliedby)$ is actually analogous to this one. Since $f$ is bounded and two limits coincide, for every $\epsilon > 0$ there exists a partition $P$ of $R$ s.t.
$$U(f,P) - L(f,P) < \epsilon$$ it follows directly from there
$$\overline I_R(f) - \underline I_R(f) < \epsilon$$ since $\epsilon$ is arbitrary these two are actually equal and henceforth $f$ is integrable. Existence of necessary partitions are maybe a bit confusing but for any $P$ shown above there must be a $P_n$ containing maybe not every subdivision point but containing sufficiently small sub-intervals containing all these subdivision points.
$f$ is integrable $\iff$ For any $\epsilon > 0$ there exists a $\delta > 0$ s.t. when $\Vert P \Vert < \delta$
Since $\Vert P_n \Vert \to 0$ while $n\to \infty$
Thanks to RRL, I believe that I corrected all the mistakes it has contained.
