# NOTE: Riemann integral

Riemann integral is a good start for learning integral as the first rigorous definition of integral on an interval.

## 1. Definition

### 1.1. Part 1

Let $$f(x)$$ be a continuous function on the interval $$[a, b]$$, we can make a $$m$$ cut($$m \in \mathbb{N}$$) to get $$m+1$$ points on $$[a, b]$$, they will be a series $$\Delta = x_0, x_1, x_2, x_3, ..., x_{m-1}, x_m$$. This series follows the following rule:

$a = x_0 \le x_1 \le x_2 \le x_3 \le ... \le x_{m-1} \le x_m = b$

These points split $$[a, b]$$ into $$m$$ subintervals $$[x_{k-1}, x_k]$$, where $$k = 1, 2, 3, ..., m$$. Now, forall $$k$$ we pick a $$\xi_k$$ that $$x_{k-1} \le \xi_k \le x_k$$, make a retangle with width is $$x_k - x_{k-1}$$ and height is $$f(\xi_k)$$. Thus, we can collect all retangles' area sum via the following formula:

$\sigma\Delta = \sum_{k=1}^{m} f(\xi_k)(x_k-x_{k-1})$

By the max-min theorem, we can know there have $$M_k$$ and $$\mu_k$$ for $$f(x)$$'s max and min value on $$[x_{k-1}, x_k]$$. As above sum of retangle's area, but for $$M_k$$ and $$\mu_k$$(they are kind of $$\xi_k$$ of course), we have the following two formulas

• Maximum $$S\Delta$$

$S\Delta = \sum_{k=1}^{m} M_k(x_k-x_{k-1})$

• and Minimum $$s\Delta$$

$s\Delta = \sum_{k=1}^{m} \mu_k(x_k-x_{k-1})$

Since

$\mu_k \le f(\xi_k) \le M_k$

, we know

$s\Delta \le \sigma\Delta \le S\Delta$

### 1.2. Part 2

We know $$f(x)$$ is uniformly continuous on $$[a, b]$$. So, for any $$\epsilon \in \mathbb{R}^+$$, exists a $$\delta(\epsilon) \in \mathbb{R}^+$$, let arbitrary two points $$x$$ and $$x'$$, if the following formula is true

$ x - x' < δ(ε)$

then

$ f(x) - f(x') < ε$

is true. And since

• $$\mu_k = f(\alpha_k)$$
• $$M_k = f(\beta_k)$$
• $$x_{k-1} \le \alpha_k \le x_k$$
• $$x_{k-1} \le \beta_k \le x_k$$

If $$x_k - x_{k-1} \lt \delta(\epsilon)$$, then $$M_k - \mu_k \lt \epsilon$$. Thus, if we write the maximum value of interval's length as the follow(notice $$x_k - x_{k-1}$$ can be different for different $$k$$, since we didn't require same length for these subintervals):

$\delta[\Delta] = \max\limits_{k}(x_k - x_{k-1})$

Then once $$\delta[\Delta] \lt \delta(\epsilon)$$, we have

$S\Delta - s\Delta = \sum_{k=1}^{m} (M_k - \mu_k)(x_k-x_{k-1}) \lt \epsilon \sum_{k=1}^{m} (x_k-x_{k-1}) \lt \epsilon(b - a)$

Now, make a new $$\epsilon' = \epsilon/(b - a)$$, we have $$\delta(\epsilon')$$ that if $$\delta[\Delta] \lt \delta(\epsilon')$$ then $$S\Delta - s\Delta \lt \epsilon'$$. Since $$\epsilon' \in \mathbb{R}$$, we can rewrite to $$\delta[\Delta] \lt \delta(\epsilon)$$ then $$S\Delta - s\Delta \lt \epsilon$$.

### 1.3. Part 3

We have a parition $$\Delta$$, using the same way can get another parition $$\Delta'$$, their union is another parition $$\Delta''$$. Of course, there will also have $$\sigma\Delta$$, $$\sigma\Delta'$$, and $$\sigma\Delta''$$ as sum of retangles from paritions.

Now, it's easy to find, $$\Delta'' = x_0'', x_1'', x_2'', ..., x_p''$$ can parition any $$[x_{k-1}, x_k]$$ in $$\Delta$$. Remember $$\mu_k$$ is the minimum value of $$f(x)$$ in $$[x_{k-1}, x_k]$$, so $$\mu_k \le f(\xi_p'')$$, where $$\xi_p'' \in \Delta''$$. Then mapping $$x_h'' = x_{k-1}$$ and $$x_j'' = x_k$$. Thus, we get

$\mu_k(x_k - x_{k-1}) = \sum_{p=h+1}^{j} \mu_k(x_p'' - x_{p-1}'') \le \sum_{p=h+1}^{j} \xi_p''(x_p'' - x_{p-1}'')$

We conclude

$s\Delta \le \sigma\Delta''$

Now consider $$S\Delta'$$, we know $$\sigma\Delta'' \le S\Delta'$$ via same process, conclude

$s\Delta \le \sigma\Delta'' \le S\Delta'$

Let's consider all partitions $$\Delta$$ of $$[a, b]$$, the corresponding $$s\Delta$$ of has a supremum

$s = \sup\limits_\Delta s\Delta$

Obviously, $$s \le S\Delta'$$, and since $$\Delta'$$ can be arbitrary parition so

$\forall \Delta, s\Delta \le s \le S\Delta$

Combine part 1 and part 2, conclude if $$\delta[\Delta] \lt \delta(\epsilon)$$, then $$|\sigma\Delta - s| \lt \epsilon$$.

### 1.4. Final

Now we can say for any partition, whatever how we pick $$\xi_k$$ for a partition $$\Delta$$, we have

$ ∑k=1mf(ξk)(xk - xk-1) - s < ε$

When $$\delta[\Delta] \rightarrow 0$$, say $$\sigma[\Delta] = \sum_{k=1}^{m}f(\xi_k)(x_k - x_{k-1})$$ has limit $$s$$, write as

$s = \lim_{\delta[\Delta] \to 0} \sum_{k=1}^{m}f(\xi_k)(x_k - x_{k-1})$

We call it the definite integral of $$f(x)$$ on $$[a, b]$$, represent as

$\int_{a}^{b} f(x)dx = \lim_{\delta[\Delta] \to 0} \sum_{k=1}^{m}f(\xi_k)(x_k - x_{k-1})$

## 2. Terminology

1. $$f(x)$$ is the integrand of $$\int_{a}^{b} f(x)dx$$
2. $$\int_{a}^{b} f(x)dx$$ called the integrate of $$f(x)$$ from $$a$$ to $$b$$
3. $$a$$ and $$b$$ called the lower and upper limit of the definite integral

Date: 2022-06-03 Fri 00:00