NOTE: Riemann integral

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

\[ s\Delta \le \sigma\Delta \le S\Delta \]

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

\begin{equation} |x - x'| \lt \delta(\epsilon) \end{equation}

then

\begin{equation} |f(x) - f(x')| \lt \epsilon \end{equation}

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 \(S\Delta - s\Delta \lt \epsilon\).

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 \(|\sigma\Delta - s| \lt \epsilon\).

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

\begin{equation} |\sum_{k=1}^{m}f(\xi_k)(x_k - x_{k-1}) - s| \lt \epsilon \end{equation}

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

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