I'm trying to show that one definition implies the other for Darboux Integrability. But, I don't really how to proceed.

Here's the first statement.

For all $\epsilon > 0$, there exists a partition P of $[a,b]$ such that $\vert U_p (f) - L_p (f) \vert < \epsilon$.

Here's the second statement.

For all $\epsilon > 0$, there exists a $\delta > 0$ such that $mesh (P) < \delta$ implies $\vert U_p (f) - L_p (f) \vert < \epsilon $.

How do I prove something like this? Any help? What's the strategy?

  • 1
    $\begingroup$ The second statement clearly implies the first one. $\endgroup$ – Berci Mar 30 '13 at 1:45

The two statements are equivalent, but some work is needed to prove (i)$\Rightarrow$(ii).

For any interval $Q$ write $\sup_{x\in Q} f(x)-\inf_{x\in Q}f(x)=:|\Delta f|_Q$. If $P$ is a partition of $[a,b]$ into $N$ subintervals $Q_k$ then $$U_P(f)-L_P(f)=\sum_{k=1}^N |\Delta f|_{Q_k}\>|Q_k|=: D_P(f)\ .$$ We need the following

Lemma. If $P$ is a partition of $[a,b]$ into $N$ subintervals $Q_k$ of length $|Q_k|>0$, and if $P'$ is another partition of $[a,b]$ satisfying the condition $${\rm mesh}(P')\leq\min_{1\leq k\leq N} |Q_k|$$ then $$D_{P'}(f)\leq 3 D_P(f)\ .$$

Granting the Lemma for the moment we can argue as follows: Assume (i). Given an $\epsilon>0$, we can find a partition $P$ of $[a,b]$ into subintervals $Q_k$ such that $D_P(f)<{\epsilon\over3}$. Let $\delta:=\min_{1\leq k\leq N}|Q_k|$. Then by the Lemma for any partition $P'$ with mesh size $\leq\delta$ we have $D_{P'}(f)<\epsilon$. This proves (ii).

Proof of the Lemma. Denote the subintervals of $P'$ by $Q_j'$ and define $$\epsilon_{jk}:=\cases{1\quad&$Q_j\cap Q_k'\ne\emptyset$\cr 0&(else)}\ .$$ Then for each $j$ we have $$|\Delta f|_{Q_j'}\leq\sum_k\epsilon_{jk}|\Delta f|_{Q_k}\ ,$$ and for each $k$ we have $$\sum_j \epsilon_{jk}|Q_j'|\leq |Q_k|+2\max_l|Q_l'|\leq|Q_k|+2\min_l |Q_l|\leq 3|Q_k|\ .$$ It follows that $$D_{P'}(f)=\sum_j|\Delta f|_{Q_j'}|Q_j'|\leq\sum_{j,\ k}\epsilon_{jk}|\Delta f|_{Q_k}|Q_j'|\leq 3\sum_k|\Delta f|_{Q_k}|Q_k|=3D_P(f)\ .$$

  • $\begingroup$ Nice proof: two interesting points. (1) $\sum_j \epsilon_{jk}|Q_j'|\leq |Q_k|+2\max_l|Q_l'|$ says the total length of sub-intervals in $P'$ which can intersect a sub-interval in $P$ is at most the length of that sub-interval in $P$ plus a sub-interval of Q overlapping at both ends. (2) If trying to work separately with the sup and inf, the proof fails since it relies on $|\Delta f|_{Q_k} \ge 0 $ to move from $|\Delta f|_{Q_j'}\leq max(\epsilon_{jk}|\Delta f|_{Q_k}\ )$ to $|\Delta f|_{Q_j'}\leq\sum_k\epsilon_{jk}|\Delta f|_{Q_k}\ $. $\endgroup$ – Tom Collinge Sep 9 '18 at 15:15
  • $\begingroup$ I think the definition of $\epsilon_{jk}$ is the wrong way around. Shouldn't it be $\epsilon_{jk}:=\cases{1\quad&$Q_j'\cap Q_k\ne\emptyset$\cr 0&(else)}\ $? $\endgroup$ – DerivativesGuy Feb 3 '20 at 18:19

The second statement is stronger by imposing an extra condition about the mesh size.So the

second statement can't be true without first.In other words, 2 implies 1.

But this line of thinking depends on the very fact that

" For every $\delta$ there exist a P with mesh size < $\delta$"

which should be trivial.

And for 1 implies 2 , you cleverly choose $\delta$ depending on the parameters of an existing

partition P satisfying 1.

You may look at the full solution here : Darboux Integrability epsilon-delta proof


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