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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.

tchappy ha
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    It is a sequence of partitions that are increasing in fineness. It follows that the $L$'s ( $U$'s) are increasing ( decreasing), so clearly convergent. It is not obvious that it will converge to the lower ( upper) Darboux integral. Rating: lower overhead cost that will require extra work in the future. An artifice I would say. – orangeskid Feb 13 '23 at 06:15
  • @orangeskid Thank you very much for your comment. – tchappy ha Feb 13 '23 at 06:33

1 Answers1

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$2^n$ 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.

Arthur
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