Do I need to prove $\lim_{\|P\|\to 0} L(f,P)=\sup_P L(f,P)$, and $\lim_{\|P\|\rightarrow 0} U(f,P)=\inf_P U(f,P)$? Or is it trivial? I'm trying to show that $f$ is bounded and Riemann integrable on $[a, b]$ implies that there exists a sequence $\{P_1, P_2, P_3, \ldots\}$ of partitions of $[a, b]$ s.t.
$$
\lim_{n\to \infty} [U_f(P_n)-L_f(P_n)] = 0
$$
in which case
$$
\int\limits_a^b f(x) dx = \lim_{n\to \infty} U_f(P_n) = \lim_{n\to \infty} L_f(P_n).
$$
(This is one part of an iff proof of the sequential characterization of the Riemann integral)
I wanted to do a proof by contradiction, and I found something I think would help finish my proof from this answer, but I'm not sure if I need to prove this or if it's some kind of definition or something. I've looked everywhere and can't find anything.
I'll lay out what I have so far:

Suppose $\lim_{n\to \infty} [U_f(P_n)-L_f(P_n)] \not= 0$. Then $\lim_{n\to \infty} U_f(P_n) - \lim_{n \to \infty}L_f(P_n) \not= 0$
$\Rightarrow \lim_{n\to \infty} U_f(P_n) \not= \lim_{n \to \infty}L_f(P_n)$.
Observe that for $\|P\|$ defined as the length of the largest subinterval $[x_{i-1}, x_i]$, $1 \leq i \leq n$, as $n \rightarrow \infty, \|P\| \rightarrow 0$ since the more partitions there are over the fixed interval, the smaller they must be in length in order to fit into the interval.
Then $\lim_{\|P\|\to 0} U_f(P_n) \not= \lim_{\|P\| \to 0}L_f(P_n)$ by the above explanation.

This is where I'm stuck, because if I can just say that $\lim_{\|P\|\to 0} L(f,P)=\sup_P L(f,P)$, and $\lim_{\|P\|\rightarrow 0} U(f,P)=\inf_P U(f,P)$ then we know the upper and lower Darboux integrals aren't equal, and thus $f$ isn't Riemann integrable, which is a contradiction.
Any help is appreciated, thanks in advance.
 A: If $f$ is Riemann integrable, then upper and lower Darboux integrals are equal and
$$\sup_P L(P,f) = \underline{\int_a}^b f(x) \, dx = \underbrace{\int_a^b f(x) \, dx}_I= \overline{\int_a^b} f(x) \, dx =\inf_P U(P,f)$$
For any $n \in \mathbb{N}$, by basic properties of $\sup$ and $\inf$, there exist partitions $P_n'$ and $P_n''$ such that
$$I-\frac{1}{n} < L(P'_n,f) \leqslant I \leqslant U(P''_n,f)< I+\frac{1}{n}$$
Let $P_n = P_n'\cup P_n''$ be the common refinement.  Since lower sums increase and upper sums decrease as the partition is refined, we have
$$I-\frac{1}{n} <L(P'_n,f) \leqslant L(P_n,f) \leqslant U(P_n,f) \leqslant U(P_n'',f) < I + \frac{1}{n}$$
Thus, $ 0 \leqslant U(P_n,f)- L(P_n,f) < \frac{2}{n}$ for all $n$ which implies that
$$\lim_{n \to \infty}[U(P_n,f) - L(P_n,f)] = 0.$$

Proof that $\lim_{\|P\| \to 0}U(P,f) = \inf_P U(P,f)$ and $\lim_{\|P\| \to 0}L(P,f) = \sup_P L(P,f)$.
Suppose $f:[a,b] \to \mathbb{R}$ is bounded and $|f(x)|\leqslant B$ for all $x \in [a,b]$.
Let $\bar{I} = \inf_PU(P,f)$.  For any $\epsilon > 0$ there exists  a partition $P_\epsilon=(x_0,x_1,\ldots,x_m)$ such that $\bar{I}\leqslant U(P_\epsilon,f) < \bar{I} + \frac{\epsilon}{2}$.
Note that $P_\epsilon$ has $m$ subintervals. Let $\delta = \frac{\epsilon}{4mB}$ and let $P$ be any partition with $\|P \| < \delta$.   The common refinement  $Q = P \cup P_\epsilon$ must have at most $m-1$ more points than $P$. (The largest discrepancy in the number of points occurs when each of the points $x_1,\ldots, x_{m-1}$ of $P_\epsilon$ is in the interior of a subinterval of $P$.)
Since $Q$ is a refinement of $P_\epsilon$, we have
$$\tag{1}\bar{I}\leqslant U(Q,f) \leqslant U(P_\epsilon,f) < \bar{I} + \frac{\epsilon}{2}$$
We claim that the difference in the upper sums for $P$ and $Q$ is bounded as
$$\tag{2}U(P,f) - U(Q,f) < (m-1) \cdot 2B \cdot \delta  = (m-1) \cdot 2B \cdot \frac{\epsilon}{4mB} < \frac{\epsilon}{2}$$
To prove the claim suppose first that $Q$ has only one more point $y^*$ in the interior of the subinterval $(y_{j-1},y_j)$ of partition $P$.  It follows that
$$U(P,f) - U(Q,f) \\= \sup_{x \in [y_{j-1},y_j]} f(x)(y_j- y_{j-1}) - \sup_{x \in [y_{j-1},y^*]} f(x)(y^*- y_{j-1})- \sup_{x \in [y^*,y_j]} f(x)(y_j- y^*) \\ \leqslant 2B \cdot (y_j - y_{j-1}) \leqslant 2B \|P\| < 2B\delta =  2B \frac{\epsilon}{4mB} = \frac{\epsilon}{2m}$$
By induction as we add up to $m-1$ additional points to $Q$ we get the claimed result (2).
Thus, $U(P,f) < U(Q,f) + \frac{\epsilon}{2}$ and using (1), we have with $\|P\| < \delta$,
$$\bar{I} \leqslant U(P,f) < U(Q,f) + \frac{\epsilon}{2} < \bar{I} + \epsilon$$
This proves that $\lim_{\|P\| \to 0} U(P,f) = \bar{I}= \inf_PU(P,f)$.
In a similar way we can show that $\lim_{\|P\| \to 0} L(P,f) = \underline{I}= \sup_P L(P,f)$.
