[Math] Darboux Integrability epsilon-delta proof

Question

I'm having trouble proving this theorem.

Suppose F is Darboux integrable on I, then for all $\epsilon > 0$, there exists a $\delta > 0$ such that mesh (P) < $\delta$ implies $\vert U_p (f) – L_p (f)\vert < \epsilon$.

I have tried to prove it by letting $
\delta = \frac{\epsilon}{\sum_{k=1}^{n}(M_k – m_k)}
$ ,where $M_k$ and $m_k$ are the suprema and infima of the kth sub-interval.

I've also chosen delta to be $\delta = \frac{\epsilon}{n(\sup I – \inf I)}$.

One can easily show that any of those two choices of delta result in $\vert U_p (f) – L_p (f)\vert < \epsilon$. However, both of those choices depend on a partition that was selected a priori (but we don't know which partition), so the proof is ultimately incorrect.

My question is how do I go about tackling this problem? Should I consider doing a proof by contradiction? what would be a better choice for delta? I think I need a hint that sets me back on the right track.

Thank you, everyone.

Answer 1

A slightly more general result is proved in $\S$ 8.4.2 of my honors calculus notes:

Dicing Lemma: Let $f: [a,b] \rightarrow \mathbb{R}$ be a bounded function. Then for all $\epsilon > 0$ there is a $\delta > 0$ such that if $\mathcal{P}$ is a partition of $[a,b]$ with mesh less less than $\delta$, then $$\underline{\int_a^b} f - L(f,\mathcal{P}) < \epsilon \text{ and }\ U(f,\mathcal{P}) - \overline{\int_a^b f} < \epsilon.$$

I call it the Dicing Lemma, and I took the exposition from online notes of D. Levermore.