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.

  • $\begingroup$ How would that solve the problem? I'm actually trying to get rid of the N, because N depends on the partition. It is not very clear. Please, explain. $\endgroup$ – jll90 Mar 28 '13 at 3:32
  • $\begingroup$ I think this question deserves more attention..But people generally don't like these stuff.. $\endgroup$ – Halil Duru Apr 2 '13 at 11:39

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. $\hspace{164mm}$ $\blacksquare$

  • 1
    $\begingroup$ I assume that by M and m you mean the supremum and infimum on that particular interval right? $\endgroup$ – jll90 Apr 2 '13 at 0:43
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    $\begingroup$ and why do we need to consider P_min? $\endgroup$ – jll90 Apr 2 '13 at 0:55
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    $\begingroup$ finally, by overlapping does that mean that a certain subinterval of the partition D contains one of the endpoints of a subinterval in P? $\endgroup$ – jll90 Apr 2 '13 at 1:13
  • $\begingroup$ M and m are global max. and min. P_min is required to guarantee that the subintervals of our new partition D doesn't contain parts of more than 2 subintervals of P.overlapping :a subinterval of D that contains one of the endpoints of a subinterval in P as an "interior" point. $\endgroup$ – Halil Duru Apr 2 '13 at 7:09
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    $\begingroup$ I have filled in the details. Your answer has proven most useful. $\endgroup$ – jll90 Apr 4 '13 at 4:40

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.

  • $\begingroup$ Thanks. I will look into this. $\endgroup$ – jll90 Apr 2 '13 at 16:26
  • $\begingroup$ along the same lines ..+1 $\endgroup$ – Halil Duru Apr 3 '13 at 18:12
  • $\begingroup$ and a good name for the lemma---good to know.. $\endgroup$ – Halil Duru Apr 3 '13 at 18:13

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