TITLE: Darboux integrability implies Riemann integrability
QUESTION [8 upvotes]: I have searched the site for posts regarding Darboux integrability $\implies$ Riemann integrability, but haven't found any that specifically adress this question.
My definition of Darboux integrability: Let $f$ be defined and bounded on $[a,b]$, then $f$ is Darboux integrable if for all $\epsilon >0$ there exists a partition $P$ of $[a,b]$ such that $U(f,P)-L(f,P)<\epsilon$ (where $U$ and $L$ are the upper and lower Riemann sums respectively).
My definition of Riemann integrability: Let $f$ be defined and bounded on $[a,b]$, then $f$ is Riemann integrable if $$\lim_{N\to\infty} \sum\limits_{k=1}^{N} f(c_k)(x_{k}-x_{k-1})$$ has the same limit for all sequences of partitions $P_N$ and all choices of $c_k\in[x_{k-1},x_{k}]$.
If my definitions are correct, it seems that Darboux integrability only requires one partition to fulfil the epsilon-inequality, whereas Riemann integrability requires all sequences of partitions to be fulfilled. How can this lead to an implication nevertheless?

REPLY [2 votes]: Hint
One has anyway $$\lim_{\|P\|\to 0} L(f,P)=\sup_P L(f,P)$$ and $$\lim_{\|P\|\rightarrow 0} U(f,P)=\inf_P U(f,P)$$ If for all $\varepsilon >0$ there exists a partition $P$ such that $U(f,P)-L(f,P)<\varepsilon$, those limits have the same value.
Then $$\lim_{\|P\|\to 0} S(f,P)$$ exists and is equal to that value since $$L(f,P) \le S(f,P) \le U(f,P)$$ for every, however tagged, partition $P$.
The sequential version of the last limit is valid too but one has to suppose $\,\|P_N\| \to 0\,$ of course.