I am reading "Measure, Integration & Real Analysis" by Sheldon Axler.
The following exercise is Exercise 7 in Exercises 1A on p.8.
Suppose $f:[a,b]\to\mathbb{R}$ is a bounded function. For $n\in\mathbb{Z}^+$, let $P_n$ denote the partition that divides $[a,b]$ into $2^n$ intervals of equal size. Prove that $$L(f,[a,b])=\lim_{n\to\infty} L(f,P_n,[a,b]) \text{ and }U(f,[a,b])=\lim_{n\to\infty} U(f,P_n,[a,b]).$$
The author divided $[a,b]$ into $2^n$ intervals of equal size in the above exercise.
The author didn't write as follows:
Suppose $f:[a,b]\to\mathbb{R}$ is a bounded function. For $n\in\mathbb{Z}^+$, let $P_n$ denote the partition that divides $[a,b]$ into $n$ intervals of equal size. Prove that $$L(f,[a,b])=\lim_{n\to\infty} L(f,P_n,[a,b]) \text{ and }U(f,[a,b])=\lim_{n\to\infty} U(f,P_n,[a,b]).$$
Suppose $f:[a,b]\to\mathbb{R}$ is a bounded function. For $n\in\mathbb{Z}^+$, let $P_n$ denote the partition that divides $[a,b]$ into $3^n$ intervals of equal size. Prove that $$L(f,[a,b])=\lim_{n\to\infty} L(f,P_n,[a,b]) \text{ and }U(f,[a,b])=\lim_{n\to\infty} U(f,P_n,[a,b]).$$
Why $2^n$ intervals?
Any reasonable reason?
The following exercise is Exercise 8 in Exercises 1A on p.8.
Suppose $f:[a,b]\to\mathbb{R}$ is Riemann integrable. Prove that $$\int_{a}^{b} f = \lim_{n\to\infty} \frac{b-a}{n}\sum_{j=1}^{n} f(a+\frac{j(b-a)}{n}).$$
In this exercise, the author divided $[a,b]$ into $n$ intervals of equal size.
$2^n$ intervals is the simplest option that guarantees that the sequences $L(f, P_n)$ and $U(f, P_n)$ are monotonous. From one $n$ to the next, you necessarily use all the same partition points, along with a bunch of new ones. This isn't the case if you use for instance $n$ partitions instead.
Using $3^n$ would also work, but $2^n$ is marginally simpler. Enough so that if the author has used $3^n$ instead, I would be truly curious as to why they made such a choice.