One epsilon-delta statement implies the other. Darboux Integrability. 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 $.
How do I prove something like this? 
Any help? What's the strategy? 
 A: The two statements are equivalent, but some work is needed to prove (i)$\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
$$D_{P'}(f)=\sum_j|\Delta f|_{Q_j'}|Q_j'|\leq\sum_{j,\ k}\epsilon_{jk}|\Delta f|_{Q_k}|Q_j'|\leq 3\sum_k|\Delta f|_{Q_k}|Q_k|=3D_P(f)\ .$$
A: 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
