Different ways of constructing a partition of an interval

Question

Let $a<b\in\mathbb{R}$. A sequence $P:=(p_0,\ldots,p_n)$ is a called a partition of $[a,b]$ if $$a=p_0<\ldots<p_n=b.$$ The size of $P$ is taken to be $\max_i(p_{i+1}-p_i)$.

Now, suppose we are given $\delta>0$. There exists $n\in\mathbb{N}_{\geq 1}$ such that $(b-a)/n<\delta$. I can divide $[a,b]$ into $n$ equal sub-intervals, by writing $$I_k:=\left[a+\frac{k}{n}(b-a),a+\frac{k+1}{n}(b-a)\right]$$ for all $0\leq k\leq n-1$. Clearly the length of each $I_k$ is $<\delta$. We take $p_k=a+\frac{k}{n}(b-a)$ for $0\leq k\leq n$.

The above gives us a partition. Are there other ways of constructing partitions of $[a,b]$ whose size is $<\delta$?

Answer 1

Here is my favorite:

LEMMA. [Cousin partitioning lemma] Let $\delta(x)$ be a positive function defined on some fixed interval $[a,b]$. Then for any subinterval $[c,d]\subset [a,b]$ there must exist points $$c=x_0<x_1< x_2< \dots < x_k = d$$ and points $\xi_i\in [x_{i-1},x_i]$ subject to the constraint that $(x_{i}-x_{i-1})< \delta(\xi_i)$ for each $i=1,2,\dots, k$.

Cousin's lemma first appeared in an 1895 paper by the Belgian mathematician Pierre Cousin who was a student of Poincaré. It was discovered again by Goursat who included it a paper that appeared in the very first issue of the American Math. Society Transactions journal in 1900.

It is very useful. Think of it as equivalent to the nested interval property. Anything you have proved before with the nested interval property (or the Bolzano-Weierstrass theorem, or the Heine-Borel theorem, or the least upper bound propery) usually has a simpler proof using Cousin's lemma.

If you use partitions of small size $\delta>0$ as you mentioned, you can define the Riemann integral. If you use Cousin's partitions monitored by a small positive function $\delta(x)$ you can define the Lebesgue integral.

Answer 2

You gave an example of a uniform partition.

The dyadic partition $P_n = (x_0,x_1,\ldots,x_{2^n})$ with $(b-a)/2^n < \delta$ and

$$x_k = a + \frac{b-a}{2^n}k, \quad k=0,1,\ldots, 2^n,$$

is also uniform and is useful because it forms a sequence of refining partitions. That is, $P_{n+1} \supset P_n$ for all $n \geqslant 1$ -- which does not hold for your sequence of uniform partitions.

There is an endless variety of other possibilities.

Answer 3

Geometric partition for integrating $f(x) = \frac{1}{x}$ on interval $[1, b]$ for defining the natural logarithm $\,\ln b$: $$ F(b) = \int_1^b \frac{1}{x} \, dx. $$

Fix $r>1$, and define $$ x_k = r^k \quad (0 \leq k \leq n). $$

The width of $k$th subinterval is $$ \Delta x_k := x_k - x_{k-1} = r^k - r^{k-1} = r^{k-1}(r-1), $$ which for a given $r$ increases as a function of $k$. Thus the size of the partition is the width of the final subinterval, i.e. $$ \Delta x_n = r^{n-1}(r-1), $$ which can be made as small as we like by shrinking the geometric ratio $r$ and increasing the number $n$.

Here's the rub: the area of each rectangle in the Riemann sum is a constant! This is the calculation for the left-hand sums (sampling on each subinterval $[x_{k-1}, x_k]$ at the left endpoint $x_{k-1}$): $$ A_k = f(x_{k-1}) \, \Delta x_k = \frac{1}{r^{k-1}} \, r^{k-1}(r-1) = r-1 $$ so the left-hand sum (an overestimate since $f$ is decreasing) is $$ L_n = \sum_{k=1}^n A_k = \sum_{k=1}^n (r-1) = n(r-1). $$ But what is $n$? Choose the least integer $n$ such that $x_n \geq b$ for the upper limit of the sum: $$ r^n = b \quad\Longrightarrow\quad n = \biggl\lceil \frac{\ln b}{\ln r} \biggr\rceil $$ Thus the left-hand sum is $$ \biggl\lceil \frac{\ln b}{\ln r} \biggr\rceil \, (r-1) \approx \frac{(r-1)}{\ln r} \, \ln b \quad\to\quad \ln b $$ as $r \to 1^+$ (equivalently, $n \to \infty$).

With a little care, this can be made rigorous. A very similar calculation, using right-hand sums and choosing $n$ via a floor function, gives an underestimate for the definite integral. Thus, by the Squeeze Theorem, we have a direct proof that $$ \int_1^b \frac{1}{x} \, dx = \ln b $$ that does not rely on any of the calculus of the natural exponential function. In fact, we can derive all the derivative and integral formulas for the exponential function from this!

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Here's an interactive graph where you can drag $r \to 1^+$ and witness $L_n \to \ln b$, visually.