Let \( f : \left[ a, \ b \right] \rightarrow \mathbb{R} \) be a bounded function. \( f \) is integrable on \( \left[ a, \ b \right] \) if and only if for any \( \epsilon > 0 \) there exists some partition \( R \) of \( \left[ a, \ b \right] \) such that

$$U \! \left( f, \ R \right) – L \! \left( f, \ R \right) < \epsilon$$

To prove this statement each of the following conditional statements must be true.

If \( f \) is integrable then for every \( \epsilon > 0 \) there exists some partition \( R \) of \( \left[ a, \ b \right] \) such that \(U \! \left( f, \ R \right) – L \! \left( f, \ R \right) < \epsilon\).

If for every \( \epsilon > 0 \) there exists some partition \( R \) of \( \left[ a, \ b \right] \) such that \(U \! \left( f, \ R \right) – L \! \left( f, \ R \right) < \epsilon\) then \(f\) is integrable.

To prove the first statement assume \(f\) is integrable. Therefore, it is the case that

$$\int_a^b f = \overline{ \int_a^b } f = \underline{ \int_a^b } f = \inf_Q U \! \left( f, \ Q \right) = \sup_P L \! \left( f, \ P \right)$$

By using the theorem of infimum and supremum respectively it is true that

$$U \! \left( f, \ Q \right) \lt \int_a^b f + \frac{ \epsilon}{2}$$
$$L \! \left( f, \ P \right) \gt \int_a^b f – \frac{ \epsilon}{2}$$

If \(R = P \cup Q\) then \(L \left( f, \ P \right) \le L \left( f, \ R \right) \le U \left( f, \ R \right) \le U \left( f, \ Q \right)\) and it is true

$$U \! \left( f, \ R \right) \le U \! \left( f, \ Q \right) \lt \int_a^b f + \frac{ \epsilon}{2}$$
$$L \! \left( f, \ R \right) \ge L \! \left( f, \ P \right) \gt \int_a^b f – \frac{ \epsilon}{2}$$

subtracting

$$U \! \left( f, \ R \right) – L \! \left( f, \ R \right) \lt \int_a^b f + \frac{ \epsilon}{2} – \left( \int_a^b f – \frac{ \epsilon}{2} \right) = \epsilon$$

To prove the second statement assume \( \epsilon \gt 0 \) and \( U \! \left( f, \ R \right) – L \! \left( f, \ R \right) \lt \epsilon \). Moreover, by definition

$$\underline{ \int_a^b } f \le \overline{ \int_a^b } f$$
$$U \! \left( f, \ R \right) \ge \overline{ \int_a^b } f$$
$$L \! \left( f, \ R \right) \le \underline{ \int_a^b } f$$

subtracting

$$0 \le \overline{ \int_a^b } f – \underline{ \int_a^b } f \le U \! \left( f, \ R \right) – L \! \left( f, \ R \right) \lt \epsilon$$

Since \( \epsilon \) is arbitrary it is the case that

$$\overline{ \int_a^b } f = \underline{ \int_a^b } f$$

thus, \( f \) is integrable.