Corollary : Let $I=[a,b]$ and let $f$ be bounded on $I$. If $\{P_n : n \in \mathbb{N} \}$ is a sequence of partitions of $I$ such that $\lim_{n \to \infty} U(f;P_n)-L(f;P_n)=0$ then $f$ is integrable and $\lim_{n \to \infty} U(f;P_n) = \int_a^b f(x) \, dx = \lim_{n \to \infty} L(f;P_n)$.
In the given result, I need to understand if i replaced Darboux sums ( upper and lower sums) with Riemann sums in the limit, will the result still hold? and why isn't the converse true? even if i took $S(f;P_n)-S(f;Q_n)$ and as the difference went to zero, would it be an Riemann integrable function? I think the converse of this result should be true,saying f is integrable allows me to take any difference of sums ( either Riemann or Darboux) and will converge to zero as the mesh of $P_n$ goes to zero. Also if i took Riemann sums instead in the limit for a sequence of partitions, satisfying the condition, then f still is integrable.
What does work is this. Let $f$ be bounded on $I = [0,1]$. Let $P_n$ be a sequence of partitions of $I$. Let $L \in \mathbb R$.
Assume: for every choice $Q_n$ of tags for $P_n$ we have $S(P_n,Q_n,f) \to L$. Then $f$ is Riemann integrable and $\int_a^b f(x)\;dx = L$.
Explanation of notation: If $P = \{a=x_0 < x_1 < \dots < x_k = b\}$ is a partition, then a choice of tags for $P$ is a set $Q = \{t_1,t_2,\dots,t_k\}$ such that $x_{j-1} \le t_j \le x_j$ for all $j$. Then the corresponding Riemann sum is $$ S(P,Q,f) := \sum_{j=1}^k f(t_j)\;(x_j-x_{j-1}) $$
So the point is: for this direction, we must allow all choices of tags. In the proof that it works, we choose tags with $f(t_j)$ close to $\sup\{f(x) : x_{j-1} \le x \le x_j\}$ on the one hand, and close to $\inf\{f(x) : x_{j-1} \le x \le x_j\}$ on the other hand.
For the other direction: if $f$ is Riemann integrable, then for any choice $P_n$ of partitions with norm going to zero, we can take any choice $Q_n$ of tags we like, and conclude $S(P_n,Q_n,f) \to L$. Convenient choices for $Q$ are: left endpoint; right endpoint; max point; min point. Another sneaky choice of tags (useful in certain proofs) is the one we get from the mean value theorem, $$ f(t_j) = \frac{1}{x_{j}-x_{j-1}}\int_{x_{j-1}}^{x_j} f(x)\;dx. $$