Riemann Integrability defined by sequence of partitions 
Prove that a bounded function $f$ is integrable on $[a, b]$ if an only if there exists a sequence of partitions $ \left(P_{n}\right)^{\infty}_{n=1} $ satisfying
  $$ \lim_{n\to\infty} [U(f, P_n) - L(f, P_n)] = 0 $$
  and in this case $\int_{a}^{b} f = \lim_{n\to\infty} U(f, P_n) = \lim_{n\to\infty} L(f, P_n) $

In the forward direction, we know that f is bounded and we want to prove that such a sequence exists to make $f$ integrable. If a function is bounded and on a closed interval, we know that it is compact. We know that f will attain a maximum and minimum on this interval. If a function is continuous, we can say that it is integrable. Therefore, we can start by proving that this function is continuous. By the sequential definition of continuity, for any $(x_n) \subset A $ with $\lim x_n \rightarrow c$, $ f(x_n) = f(c) $.
Now, we must prove that $\lim_{n\to\infty} U(f, P_n) - \lim_{n\to\infty} L(f, P_n) = 0 $. How can I do this? Do I have the right idea for the proof so far? 
 A: $\Longrightarrow$
$f$ is integrable hence
$S = \sup\{L(f,P) : \text{P is a partition of [a,b] \}} = \inf\{U(f,P) : \text{P is a partition of [a,b] \}} = I$ 
Note that for simplicitly, we will call the above sets $S$ and $I$ later on. 
Now, since there exist elements of the above sets arbitrarily close to their supremum and infimum, this implies that there exist partitions $P_1$ and $P_2$ of $[a,b]$ such that $U(f,P_1) - L(f,P_2) < \epsilon$ for all positive $\epsilon$. But then if this holds, then the refinement $P$ containing both $P_1$ and $P_2$ satisfies the above inequality as well. 
Now, it's not too hard to define a sequence. For example, we can replace $\epsilon$ with $\frac 1n$ and define $P_n$ to be a partition such that $U(f,P_n) - L(f,P_n) < \frac 1n$. Then one can use the squeeze theorem to show that it converges to $0$. 
$\Longleftarrow$
Note that, in general,  $S \leq I$ 
Suppose $f$ isn't integrable. Then $\leq$ is replaced with $<$. Furthermore, we have $0<I-S\leq U(f,P) - L(f,P)$ for every partition $P$ from the definition of $\sup$ and $\inf$. 
However, since $U(f, P_n) - L(f, P_n) \to 0$ it can be made smaller than $I-S$ for $n$ large enough, contradiction. 
Lastly, to show that $\int_{a}^{b} f = \lim_{n\to\infty} U(f, P_n) = \lim_{n\to\infty} L(f, P_n)$ note $S = I = \int_{a}^{b} f$ and this value uniquely satisfies $L(f,P) \leq \int_{a}^{b} f \leq U(f,P)$ for every partition. 
Thus, if  $\lim_{n \to \infty} U(f,P_n) = \lim_{n \to \infty}  L(f,P_n)$ both must necessarily tend to $\int_{a}^{b} f$ as $n \to \infty$. A rigorous proof with $\epsilon-\delta$ is not difficult but my answer is long-winded as it is. 
