A property of Darboux sum I'm trying to show that:
$$\overline{I}:=\inf _P S(f;P)=\lim_{\lambda (P)\to 0}S(f;P)$$
where $P$ is a generic partition (made by $n$ points) of the interval $[a,b]$, $f$ is bounded on $[a,b]$, $S(f;P):=\sum _{i=1}^n\sup_{\Delta_i}f(x) \cdot \Delta x_i $ upper Darboux sum of $f$ on partition $P$ with mesh $\lambda (P)$.
Facts already known and proved:
a) $s(f;P_1):=\sum _{i=1}^{n_1}\inf_{\Delta_i}f(x) \cdot \Delta x_i \leq S(f;P_2)$, for all partitions $P_1$ and $P_2$.
b) $0\leq S(f;P)-S(f;\widetilde{P})\leq \omega (f;[a,b])\cdot (\Delta x_{k_1}+...+\Delta x_{k_m}) $, where $\widetilde{P} $ is a generic refinement of $P$, $ \omega (f;[a,b]):=\sup _{x',x'' \in [a,b]}|f(x')-f(x'')| $ and $\Delta x_{k_1},...,\Delta x_{k_m} $ are all the intervals of $P$ which contain points only in $\widetilde{P} $.
My attempt to demonstrate the statement:
$ \overline{I} $ is well defined thanks to a). Being an $\inf $, this implies that $\forall \epsilon >0$ there exists a partition $P_\epsilon $ such that:
$$\overline{I} \leq S(f;P_\epsilon) \leq \overline{I} + \epsilon$$
to conclude, it would be enough for me to show that any partition $ P $ with a mesh that is narrower than that of $ P_\epsilon $ leads to $S(f;P) \leq S(f;P_\epsilon) $, but all I managed to get (using also b)) was this:
$$S(f;P) \leq S(f;P_\epsilon) +\omega (f;[a,b])\cdot m\cdot \lambda (P)$$
where $m$ is the number of points in $P_{\epsilon} $ and $\lambda (P) $ is the mesh of the new partition $P$.
I can't do better.
A little help, please?
Thanks in advance.
 A: You are on the right track.  To help you finish, recall that it is to be shown for any $\epsilon > 0$ there exists a partition $P$ such that $|S(f;P) - \bar{I}| < \epsilon$ or equivalently 
$$\tag{*}\bar{I} \leqslant S(f;P) \leqslant \bar{I} + \epsilon,$$
when $\lambda(P) < \delta$.
As you showed, now using $\epsilon/2$ instead of $\epsilon$, there exists a partition $P_\epsilon$ such that
$$\overline{I} \leqslant S(f;P_\epsilon) \leqslant \overline{I} + \frac{\epsilon}{2}$$
We want to introduce any partition $P$,  not necessarily a refinement of $P_\epsilon$, and produce $\delta$ such that (*) is satisfied if $\lambda(P) < \delta$. Next, we construct a partition $\widetilde{P}$ that is a common refinement of $P$ and $P_\epsilon$ by adding the $m$ interior points of $P_\epsilon$ to $P$. Now, as you came close to reasoning, it follows that
$$S(f;P) \leqslant S(f;\widetilde{P}) + \omega (f;[a,b])\cdot m\cdot \lambda (P)$$
Further insight into why this is true can be obtained either by drawing a picture or following my argument here.
Since $\widetilde{P}$ is a refinement of $P_\epsilon$, we have $S(f;\widetilde{P}) \leqslant S(f;P_\epsilon)$, and it follows that 
$$S(f;P) \leqslant S(f;P_\epsilon) + \omega (f;[a,b])\cdot m\cdot \lambda (P) \leqslant \overline{I} + \frac{\epsilon}{2}+ \omega (f;[a,b])\cdot m\cdot \lambda (P)$$
By choosing $\delta = \epsilon/ (2 \cdot m \cdot \omega (f;[a,b]))$, it follows that if $\lambda(P) < \delta$, then 
$$\bar{I} \leqslant S(f;P) \leqslant \bar{I} + \epsilon$$
