Riemann integral

You may have learned that the integral is the inverse operation of differentiation. Here, we define the integral as the calculation of area. This approach has an advantage that can be easily extended to higher dimensions.

Definition (Partition of an interval)

The partition of the closed interval \([a,b]\) is a finite sequence \(\Delta = \{x_n\}\) of the form

\[a = x_0 < x_1 < x_2 < \cdots < x_{n-1} < x_{n} = b.\]

Each \([x_{i},x_{i+1}]\) is called a sub-interval. The mesh or norm of a partition is defined to be the maximum length of the sub-intervals:

\[\max\{(x_{i+1} - x_i) \mid {i=0, 1,\cdots, n-1}\}.\]

Let \(f(x)\) be a bounded function on \([a,b]\). Let us define the following quantities:

\[ \begin{eqnarray} M_i &=& \sup\{f(x)\mid x_{i} \leq x \leq x_{i+1}\},\\ m_i &=& \inf\{f(x)\mid x_{i} \leq x \leq x_{i+1}\}. \end{eqnarray} \]

Remark. If \(f(x)\) is continuous on \([a,b]\), \(f(x)\) has maximum and minimum values on each \([x_i, x_{i+1}]\) (the Extreme Value Theorem). Thus, \(M_i\) and \(m_i\) are the maximum and minimum values. If \(f(x)\) is not continuous, it may not have maximum or minimum values. Nevertheless, since we are assuming \(f(x)\) is bounded, supremum and infimum do exist. Thus \(M_i\) and \(m_i\) are well-defined. □

See also: Continuity of a function for the extreme value theorem.

Now, 

\[\begin{eqnarray} S_{\Delta} &=& \sum_{i=0}^{n-1}M_i\cdot(x_{i+1}-x_{i}),\\ s_{\Delta} &=& \sum_{i=0}^{n-1}m_i\cdot(x_{i+1}-x_{i}) \end{eqnarray} \]

are called the upper Riemann sum and lower Riemann sum, respectively, with respect to the partition \(\Delta\).

Remark. Upper or lower Riemann sums are also known as upper and lower Darboux sums. □

Each \(m_i\cdot(x_{i+1}-x_{i})\) (or \(M_i\cdot(x_{i+1}-x_{i})\)) represents the area of a rectangle with the ``width'' of \(x_{i+1} - x_{i}\) and the ``height'' of \(m_i\) (or \(M_i\)). Note that \(m_i\) (or \(M_i\)) can be negative. So this area is a signed area.

The Riemann sums approximate the signed area enclosed by the graph of \(y=f(x)\), the \(x\)-axis, \(x = a\) and \(x = b\). Clearly,

\[s_{\Delta} \leq S_{\Delta}.\]

Definition (Refinement of a partition)

Let \(\Delta = \{x_0, x_1, \cdots, x_n\}\) and \(\Delta' = \{x_0',x_1',\cdots, x_m'\}\) be partitions of \([a,b]\). \(\Delta'\) is said to be a refinement of \(\Delta\) if each \(a_i\in\Delta\) is equal to some \(a_j'\in \Delta'\).

Remark. If \(x_i = x_j'\) and \(x_{i+1} = x_{j+k}'\), then the sub-interval \([x_{i}, x_{i+1}]\) is partitioned into smaller sub-intervals

\[[x_j',x_{j+1}'], [x_{j+1}',x_{j+2}'], \cdots, [x_{j+k-1}',x_{j+k}'].\]

□

Example. Consider the interval \(I=[0, 1]\).

• \(\Delta = \{0, 0.5, 1\}\) is a partition of \(I\). 
• \(\Delta' = \{0, 0.2, 0.5, 0.7, 0.9, 1\}\) is another partition of \(I\) and also a refinement of \(\Delta\).

□

Lemma

Let \(f(x)\) be a bounded function on \([a,b]\). Let \(\Delta\) be a partition of \([a,b]\) and \(\Delta'\) be a refinement of \(\Delta\). Then

\[\begin{eqnarray} s_{\Delta} &\leq& s_{\Delta'},\\ S_{\Delta} &\geq& S_{\Delta'} \end{eqnarray} \]

where \(s_{\Delta}\) and \(s_{\Delta'}\) are lower Riemann sums of \(f(x)\) with respect to \(\Delta\) and \(\Delta'\), respectively, and \(S_{\Delta}\) and \(S_{\Delta'}\) are the corresponding upper Riemann sums.

Proof. Let \(I_i = [x_{i}, x_{i+1}]\) be a sub-interval of \(\Delta\), and \(I_j' = [x_{j}', x_{j+1}']\) be a sub-interval of \(\Delta'\) such that \(I_j'\subset I_i\). 

Let \(m_i = \inf\{f(x)\mid x \in I_i\}\) and \(m_j' = \inf\{f(x)\mid x \in I_j' \}\).

Then, \[m_i \leq m_j'\] because \(m_i\) is at least as low as \(m_j'\) (since \(I_i\) covers a wider range than \(I_j'\)).

If \(x_i = x_j'\) and \(x_{i+1} = x_{j+k}'\), then

\[ \begin{eqnarray*} m_i\cdot(x_{i+1} - x_{i}) &=& m_i\cdot\sum_{l=0}^{k-1}(x_{j+l+1}' - x_{j+l}')\\ &=& \sum_{l=0}^{k-1}m_{i}\cdot(x_{j+l+1}' - x_{j+l}')\\ &\leq& \sum_{l=0}^{k-1}m_{j+l}'\cdot(x_{j+l+1}' - x_{j+l}'). \end{eqnarray*} \]

Summing both sides over \(i=0, 1, \cdots, n-1\), we have

\[s_{\Delta} \leq s_{\Delta'}.\]

\(S_{\Delta} \geq S_{\Delta'}\) is similarly proved. ■

If we consider a sequence of finer partitions \(\Delta, \Delta', \Delta'', \cdots\), we have 

\[s_{\Delta} \leq s_{\Delta'} \leq s_{\Delta''} \leq \cdots \leq S_{\Delta''} \leq S_{\Delta'} \leq S_{\Delta}\]

Therefore the sequence of the lower Riemann sums is monotone increasing with respect to refinements of partitions, and is bounded above by upper Riemann sums (e.g., \(S_{\Delta}\), etc). By the continuity axiom of real numbers, \(\sup_{\Delta} s_{\Delta}\) exists (\(\sup_{\Delta}\) means supremum over all possible partitions \(\Delta\)). Similarly, \(\inf_{\Delta}S_{\Delta}\) exists. Clearly,

\[\sup_{\Delta}s_{\Delta} \leq \inf_{\Delta}S_{\Delta}. \]

Based on the above observation, we define the definite integral as follows.

Definition (Definite integral)

Let \(f(x)\) be a bounded function on the closed interval \([a,b]\). Let \(s_{\Delta}\) and \(S_{\Delta}\) be the lower and upper Riemann sums of \(f(x)\) with respect to the partition \(\Delta\). The function \(f(x)\) is said to be Riemann-integrable, or simply integrable, if \(\sup_{\Delta}s_{\Delta} = \inf_{\Delta}S_{\Delta}\) and we write

\[\int_{a}^{b}f(x)dx = \sup_{\Delta}s_{\Delta} (= \inf_{\Delta}S_{\Delta})\]

which we call the definite integral of \(f(x)\) on \([a,b]\).

Example. Let us define the following function on \([0,1]\) (called the Dirichlet function):

\[ f(x) = \left\{ \begin{array}{cc} 1 & \text{if \(x\in \mathbb{Q}\)},\\ 0 & \text{if \(x \not\in \mathbb{Q}\)}. \end{array}\right. \]

This function is not Riemann-integrable. Take any sub-interval \([a, b]\) of any partition \(\Delta\). Since rational numbers are dense, there is always at least one rational number in \([a,b]\). Hence \(S_{\Delta} = 1\). Similarly, irrational numbers are dense so that \(s_{\Delta} = 0\). Therefore

\[\sup_{\Delta}s_{\Delta} = 0 < 1 = \inf_{\Delta}S_{\Delta}.\]

□