Darboux integrability implies Riemann integrability 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?
 A: Here's a direct solution in case you don't want to end up repeating the second half of Riemann-Lebesgue's proof. 
You can use the partition $P$ for which $U(f,P)-L(f,P)<\epsilon$ to argue that for all partitions $Q$ with mesh$(Q) = \|Q\|< \delta_P$, the Riemann sum will be somewhere close to $U(f, P)$ and $L(f, P)$. 
Assume that $P$ has $N$ points $\{p_1=a, p_2, \cdots, p_N=b\}$, partition $Q$ has $N'$ points $\{q_1=a, \cdots, q_{N'}=b\}$  and $|f| \leq M$ in $[a, b]$ (this should hold for some $M$ or Darboux integral won't be well-defined).
Now take $\delta_P< \min(\|p\|, \frac{\epsilon}{MN})$. Now for each $i \leq N'$, either $[q_i, q_{i+1}] \subset [p_j, p_{j+1}]$ (for some $j\leq N$), or $p_{j-1} \leq q_i \leq p_j \leq q_{i+1} \leq q_{j+1}$. The latter can happen at most $N$ times, so the area under $f$ for such cases can be at most $N \times M \times \delta_P \leq \epsilon$. For the rest, the sum happens to be nicely sandwiched by $U(f,P)$ and $L(f, P)$, proving the Riemann integrability.
A: If $U(f,P)-L(f,P)<\epsilon$, for a given partition $P$, then the inequality holds for $any$ refinement of $P$, since if $P\subseteq P'$, we have
$L(f,P)\le L(f,P')\le U(f,P')\le U(f,P).$ 
And since for any partition $P=\{a,x_1,\cdots, x_{n-2},b\}$, 
and any $\{x^*_i:x_i\le x^*_i\le x_{i+1};\ 1\le i\le n\},$
we have $L(f,P)\le \sum^{n-1}_{i=1}f(x^*_i)(x_{i+1}-x_i)\le U(f,P),$
it follows that Darboux integrability implies Riemann integrability.
edit: Assuming that the OP wants to show that if bounded function $f$ on $[a, b]$ is Darboux integrable then for each given $\epsilon> 0$, there is a $\delta > 0$ such that 
mesh$P < \delta \Rightarrow U(f, P) − L(f, P) < \epsilon,$ here is a sketch:
Suppose we have a partition $P_0=\{a = x_0 < \cdots < x_m = b\}$ such that $U(f,P_0)-L(f,P_0)<\epsilon.$ Let $P=\{a = t_0 < t_1 < \cdots < t_n = b\}$ be any other partition, fine enough so that at most one member of $P_0$ lies between any two members of $P$. 
Now, form $Q=P\cup P_0$  and note that in the difference $L(f,Q)-L(f,P)$, the only terms that remain correspond to elements of $Q$ of the form $[t_{n_k}\le x_{n_k}\le t_{n_{k+1}}]$. 
Then, with $m_k,m'_k,m''_k$ the mimima of $f$ on $[t_{n_k},x_{n_k}], [x_{n_k}, t_{n_{k+1}}]$ and $[t_{n_k}, t_{n_{k+1}}],\ $ respectively and $M$ an upper bound on $|f|$,
$L(f, Q) − L(f, P)=\sum _k[(m_k(x_{n_k}-t_{n_k})+m'_k(t_{n_{k+1}}-x_{n_k}))-m''_k(t_{n_{k+1}}-t_{n_k})]\le \sum_k[M(t_{n_{k+1}}-t_{n_k})-m_k''((t_{n_{k+1}}-t_{n_k}))]\le MK(\text{mesh P})$ where $K$ is the upper limit of the sum. 
Similarly, $U(f,P)-U(f,Q)\le MK(\text{mesh P})$.
The result now follows from the string of inequalities:
$U(f, P) − L(f, P) ≤ U(f, P) − U(f, Q) + U(f, Q) − L(f, Q) + L(f, Q) − L(f, P)$.
A: 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.
A: A function is Riemann integrable if
$$
\sup_P\{L(f,P)\}=\underline{\int_a}^b f(x) =\overline{\int_a^b} f(x)=\inf_P\{U(f,P)\},
$$
or equivalently, $\displaystyle \inf_P\{U(f,P)\}-\sup_P\{L(f,P)\}<\epsilon$ for any $\epsilon>0.$
Let $\epsilon>0.$ There exists $P$ such that $U(f,P)-L(f,P)<\epsilon.$ Then since 
$$
L(f,P)\leq \sup_P\{L(f,P)\}\leq \inf_P\{U(f,P)\}\leq U(f,P),
$$
clearly $\displaystyle \inf_P\{U(f,P)\}-\sup_P\{L(f,P)\}<\epsilon$ holds.
A: One way to see your definition of Riemann integrability is not correct is to look at $[0,1]$ and define $f$ to equal $0$ on $[0,1/2],$ $f=1$ on $(1/2,1].$ We know this function gives $\int_0^1 f =1/2$ using the Darboux definition.
But suppose $P_N=\{k/(2N): k=0,\dots,N-1\}\cup \{1\}.$ Define $c_k= k/(2N),k=1,\dots,N-1,$ $c_N=1/2.$ Then
$$\sum_{k=1}^{N}f(c_k)(x_k-x_{k-1}) = 0$$
for all $N.$
