Definitions of integrability I found two definitions of integrability.
Let $f:[a,b]\to\mathbb{R}$ be a function and $\left\{[x_{0},x_{1}],[x_{1},x_{2}],\dots ,[x_{n-1},x_{n}]\right\}$ a partition, where $a=x_{0}<x_{1}<x_{2}<\cdots <x_{n}=b$ with $x_i-x_{i-1}=\frac{b-a}{n}$ and $n\in\mathbb{N}$.

*

*$f:[a,b]\to\mathbb{R}$ is said to be Rimeann integrable on $[a,b]$ if there exists $L\in\mathbb{R}$ such that for every $\epsilon>0$ there exists $N$ such that for $n\geq N$,
$$\left\lvert\sum_{i=1}^nf(t_i)\frac{b-a}{n}-L\right\rvert<\epsilon$$
where $t_i\in [x_{i-1},x_i]$.


*$f$ is Riemann integrable if for $L\in\mathbb{R}$,  $\displaystyle\lim_{n\to\infty}\sum_{i=1}^nf(t_i)\frac{b-a}{n}=L$ for every possible choice of $t_i$.
It seems to me that Definition 2 states that every sequence in the set, $\displaystyle\left\{\left\{\sum_{i=1}^nf(t_i)\frac{b-a}{n}\right\}_{n\in\mathbb{N}}:t_i\in[x_{i-1},x_i]\right\}$ converges to $L$ and Definition 1 clearly implies Definition 2. Does Definition 2 also imply Definition 1?
 A: This is lengthy, but does eventually attempt to get at the question, albeit indirectly.
When I first took calculus, the business of taking "limits" of Riemann sums bothered me for reasons I couldn't articulate. Many years later, with more study of analysis, the reasons became clear: Limits in calculus are defined for (i) functions of one variable or (ii) real sequences. By contrast, a Riemann sum depends on a partition $P = (x_{i})_{i=0}^{n}$ of an interval and a choice of sample points $(t_{i})_{i=1}^{n}$ with $x_{i-1} \leq t_{i} \leq x_{i}$ for $1 \leq i \leq n$.
In short, an individual Riemann sum depends on an arbitrarily large finite number of variables, and taking a limit always entails letting the number of variables grow without bound (because the mesh of the partition must approach $0$).
Darboux integration of a bounded function $f$ therefore came as a huge conceptual relief: We pick a partition $P = (x_{i})_{i=0}^{n}$ of $[a, b]$, but instead of sampling the integrand $f$, we take lower and upper bounds
$$
m_{i} = \inf\{f(t) : x_{i-1} \leq t \leq x_{i}\},\qquad
M_{i} = \sup\{f(t) : x_{i-1} \leq t \leq x_{i}\},
$$
and form lower and upper sums
$$
L(f, P) = \sum_{i=1}^{n} m_{i}\, \Delta x_{i},\qquad
U(f, P) = \sum_{i=1}^{n} M_{i}\, \Delta x_{i}.
$$
Now there's no need for "limits"! Instead we take the supremum of the set of lower sums (over all partitions $P$) and the infimum of the set of upper sums, and we declare $f$ to be integrable on $[a, b]$ if these two numbers are equal.
This would be a lengthy and pointless digression in the context of this question but for an elementary observation: For every partition $P = (x_{i})_{i=0}^{n}$, for every subinterval $[x_{i-1}, x_{i}]$, and for every sample point $t_{i}$, we have
$$
m_{i} \leq f(t_{i}) \leq M_{i}.
$$
Multiplying across by $\Delta x_{i}$ and summing over $i$ shows that every Riemann sum associated to $P$ is bounded below by $L(f, P)$ and bounded above by $U(f, P)$. Particularly, if $f$ is Darboux integrable on $[a, b]$, then $f$ is Riemann integrable on $[a, b]$. The converse is also true: Loosely, by sampling suitably we can make $f(t_{i})$ as close to the infimum $m_{i}$ (or to the supremum $M_{i}$) as we like. (Usually, we like to take small to mean $< \frac{\varepsilon}{4(b-a)}$ or some such.)
What Darboux integration gives us that Riemann integration doesn't (generally) is workable lower and upper bounds for the quantity we wish to define. The moral of this story (if you ask me) is to frame questions about Riemann integrability in terms of Darboux integrability.
One final coda: In condition 2., there's an implicit existential quantifier on $L$, and presumably "Every choice of $t_{i}$" means "in the corresponding subinterval". Even with those adjustments, however, there's the vexation of the term limit, for the reasons noted above. And, there's a technical hitch in the "for every sequence in the set...", because the stated set of all Riemann sums depends only on the integrand $f$ and the interval $[a, b]$, but in order to "probe" integrability we have to restrict ourselves to sequences sampled from partitions whose mesh converges to $0$.
