If \( f : \left[ a, \ b \right] \rightarrow \mathbb{R} \) is a bounded function then for any partition \(P\) of \( \left[ a, \ b \right] \) by definition

$$m = \inf_{\left[ a, \ b \right]} f \left( x \right)$$
$$m_i = \inf_{\left[ x_{i-1}, \ x_i \right]} f \left( x \right)$$
$$M_i = \sup_{\left[ x_{i-1}, \ x_i \right]} f \left( x \right)$$
$$M = \sup_{\left[ a, \ b \right]} f \left( x \right)$$

and

$$m \le m_i \le M_i \le M$$

So by definition

$$m \left( b – a \right) \le L \left( f, \ P \right) \le U \left( f, \ P \right) \le M \left( b – a \right)$$

The above result suggests that \(P\) could be the refinement of some other partition. Consider the notion that the Lower & Upper Darboux Sums are defined using different partitions. Let

$$L \left( f, \ P \right)$$
$$U \left( f, \ Q \right)$$

Is it the case that \(L \left( f, \ P \right) \le U \left( f, \ Q \right)\)? If so then it is the case that any Upper Darboux sum is an upper bound for the set of all Lower Darboux sums. Similarly, any Lower Darboux sum will be a lower bound for the set of Upper Darboux sums.

Consider a refinement of both \( P \) and \( Q \), namely, \( R = P \cup Q \). Since any refinement \( R \) of \( P \) yields

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

Since any refinement \( R \) of \( Q \) yields

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

Therefore, it is true that

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

Moreover, the Completeness Property of the Real Numbers states that every nonempty set of real numbers bounded above must have a least upper bound. Similarly, every nonempty set of real numbers that is bounded below must have a greatest lower bound. Let

$$A = \sup_P L \left( f, \ P \right)$$

and

$$B = \inf_Q U \left( f, \ Q \right)$$

Therefore, it must be the case that

$$L \left( f, \ P \right) \le A \le B \le U \left( f, \ Q \right)$$