Darboux Integrability epsilon-delta proof 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.
 A: 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.
A: By Darboux integrability ,
There exist a partition P such that $\vert U_P(f) - L_P (f)\vert < \varepsilon$.
Let N be the number of intervals in this partition  and $P_{min}$ be the minimum length of 
the subintervals in P.
Define  $ A $ : = max {$1$,$M-m$} and  $\varepsilon$':= min {$P_{min}$,$\varepsilon$} 
Now choose $\delta$= $\varepsilon'/NA$
Let's take a partition D with this mesh $\delta$.
And observe that $|U_D(f)-L_D(f)|$< $\varepsilon$ + A  $\delta $$ N $ < $2\varepsilon $ 
where   $ A$ $\delta$ N corresponds to the contribution of the overlapping subintervals of
D in P.   $\hspace{164mm}$   $\blacksquare$
