If \( f : \left[ a, \ b \right] \rightarrow \mathbb{R} \) is a bounded function then for an arbitrary partition of \(\left[ a, \ b \right]\) let

$$M = \sup_{\left[ a, \ b \right]} f \left( x \right)$$

and

$$M_i = \sup_{\left[ x_{i-1}, \ x_i \right]} f \left( x \right)$$

The Upper Darboux sum is defined as

$$U \left( f, \ P \right) = \sum_{i=1}^n M_i \left( x_i – x_{i-1} \right)$$

Note that \(f\) needs to be defined on \(\left[ a, \ b \right]\) but it does not have to be continuous. The following diagrams illustrate possible Upper Darboux sums. The diagrams should be understood as representative but not as definitive.

graphics coming soon

Notice that if \(P\) is any partition of \( \left[ a, \ b \right] \) then \( U \left( f, \ P \right) \le M \left( b-a \right) \). By definition \( M_i \le M \) therefore,

$$\sum_{i=1}^n M_i \left( x_i – x_{i-1} \right) \le \sum_{i=1}^n M \left( x_i – x_{i-1} \right) = M \sum_{i=1}^n \left( x_i – x_{i-1} \right)$$

it is the case that

$$\sum_{i=1}^n \left( x_i – x_{i-1} \right) = \left( x_1 – x_0 \right) + \left( x_2 – x_1 \right) + \left( x_3 – x_2 \right) + \dots + \left( x_{n-1} – x_{n-2} \right) + \left( x_n – x_{n-1} \right) = \left( x_n – x_0 \right) = \left( b – a \right)$$

therefore, \( U \left( f, \ P \right) \le M \left( b-a \right) \) as claimed.

If the partition \( P \) of the above graphics are refined to a partition \( R \) then

graphics coming soon

From the above diagrams it seems reasonable to suggest \( U \left( f, \ R \right) \le U \left( f, \ P \right) \).

Consider an arbitrary interval in \( P \), \( \left[ x_{i-1}, \ x_i \right] \) refined into \( \left[ x_{i-1}, \ t \right] \cup \left[ t, \ x_i \right] \). Let

$$M_{t^{-}} = \sup_{\left[ x_{i-1}, \ t \right]} f \left( x \right)$$
$$M_{t^{+}} = \sup_{\left[ t, \ x_i \right]} f \left( x \right)$$

By definition

$$M_{t^{-}} \le M_i$$
$$M_{t^{+}} \le M_i$$

The area of the refinement is

$$M_{t^{-}} \left( t – x_{i-1} \right) + M_{t^{+}} \left( x_i – t \right) \le M_i \left( t – x_{i-1} \right) + M_i \left( x_i – t \right) = M_i \left( x_i – x_{i-1} \right)$$

Hence, reasoning inductively it is indeed true that

$$U \left( f, \ R \right) \le U \left( f, \ P \right)$$