Characterization of Riemann-Integrable Functions

A bounded function $ f(x) $ defined on the interval [a, b] is Riemann integrable if and only if, for every $ \epsilon > 0 $, there exists a partition $ P $ of the interval such that $$ S(P) - s(P) < \epsilon $$ where $ S(P) $ denotes the upper Darboux sum, $ s(P) $ the lower Darboux sum, and $ \epsilon $ is an arbitrarily small positive real number.

Proof

The proof is structured in two parts.

1] If $ s(f) = S(f) $, then $ S(f) - s(f) < \epsilon $

We begin by assuming that $ s(f) = S(f) $, and we aim to prove that $ S(f) - s(f) < \epsilon $.

The condition $ s(f) = S(f) $ implies that the function $ f(x) $ is Riemann integrable.

$$ s(f) = S(f) $$

In other words, if $ f(x) $ is Riemann integrable over [a, b], then the infimum of all upper Darboux sums coincides with the supremum of all lower Darboux sums.

Note: The value $ s(f) $ is the supremum of the set of all lower sums corresponding to various partitions (P, Q). Similarly, $ S(f) $ is the infimum of the set of upper sums over those same partitions.

Given any $ \epsilon > 0 $, we can find two partitions $ P $ and $ Q $ of [a, b] such that:

$$ s(f) - \frac{\epsilon}{2} < s(P) $$

$$ S(f) + \frac{\epsilon}{2} > S(Q) $$

Now consider their common refinement:

$$ R = P \cup Q $$

By properties of partition refinement, we have:

$$ s(f) - \frac{\epsilon}{2} < s(P) \le s(R) \le S(R) \le S(Q) < S(f) + \frac{\epsilon}{2} $$

Focusing on the refined partition $ R $, we get:

$$ s(f) - \frac{\epsilon}{2} \le s(R) \le S(R) < S(f) + \frac{\epsilon}{2} $$

Subtracting $ s(R) $ from $ S(R) $ and using $ s(f) = S(f) $:

$$ S(R) - s(R) < S(f) + \frac{\epsilon}{2} - s(f) + \frac{\epsilon}{2} $$

$$ S(R) - s(R) < \frac{\epsilon}{2} + \frac{\epsilon}{2} $$

$$ S(R) - s(R) < \epsilon $$

This shows that for every $ \epsilon > 0 $, we can find a partition $ R $ such that the difference between the upper and lower Darboux sums is less than $ \epsilon $.

2] If $ S(f) - s(f) < \epsilon $, then $ S(f) = s(f) $

This time, we assume that $ S(f) - s(f) < \epsilon $, and we want to prove that $ S(f) = s(f) $.

Consider any partition $ P $ such that:

$$ S(P) - s(P) < \epsilon $$

This doesn't necessarily imply that $ f(x) $ is Riemann integrable, since in general:

$$ S(P) \ne s(P) $$

However, we do know that:

$$ S(P) \ge S(f) $$

$$ s(P) \le s(f) $$

So we can deduce:

$$ S(f) - s(f) \le S(P) - s(P) < \epsilon $$

Now, since $ S(f) - s(f) $ is fixed and does not depend on $ \epsilon $, and $ \epsilon $ can be made arbitrarily small, the only possibility for the inequality above to hold for every $ \epsilon > 0 $ is:

$$ S(f) - s(f) = 0 $$

Which gives:

$$ S(f) = s(f) $$

This equality between the upper and lower Darboux integrals means that $ f(x) $ is Riemann integrable.

So, if there exists a partition $ P $ such that $ S(P) - s(P) < \epsilon $, then there must exist some partition where $ S(f) = s(f) $, confirming the integrability of $ f(x) $.

This concludes the second part of the proof.

And so on.