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I'm trying to show that:

$$overlineI:=inf _P S(f;P)=lim_lambda (P)to 0S(f;P)$$

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^nsup_Delta_if(x) cdot Delta x_i $ upper Darboux sum of $f$ on partition $P$ with mesh $lambda (P)$.

Facts already known and proved:

a) $s(f;P_1):=sum _i=1^n_1inf_Delta_if(x) cdot Delta x_i leq S(f;P_2)$, for all partitions $P_1$ and $P_2$.

b) $0leq S(f;P)-S(f;widetildeP)leq omega (f;[a,b])cdot (Delta x_k_1+...+Delta x_k_m) $, where $widetildeP $ 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 $widetildeP $.

My attempt to demonstrate the statement:

$ overlineI $ is well defined thanks to a). Being an $inf $, this implies that $forall epsilon >0$ there exists a partition $P_epsilon $ such that:

$$overlineI leq S(f;P_epsilon) leq overlineI + epsilon$$

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:

$$S(f;P) leq S(f;P_epsilon) +omega (f;[a,b])cdot mcdot lambda (P)$$

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.

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 $|S(f;P) - barI| < epsilon$ or equivalently

$$tag*barI leqslant S(f;P) leqslant barI + epsilon,$$

when $lambda(P) < delta$.

As you showed, now using $epsilon/2$ instead of $epsilon$, there exists a partition $P_epsilon$ such that

$$overlineI leqslant S(f;P_epsilon) leqslant overlineI + fracepsilon2$$

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 $widetildeP$ 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

$$S(f;P) leqslant S(f;widetildeP) + omega (f;[a,b])cdot mcdot lambda (P)$$

Further insight into why this is true can be obtained either by drawing a picture or following my argument here.

Since $widetildeP$ is a refinement of $P_epsilon$, we have $S(f;widetildeP) leqslant S(f;P_epsilon)$, and it follows that

$$S(f;P) leqslant S(f;P_epsilon) + omega (f;[a,b])cdot mcdot lambda (P) leqslant overlineI + fracepsilon2+ omega (f;[a,b])cdot mcdot lambda (P)$$

By choosing $delta = epsilon/ (2 cdot m cdot omega (f;[a,b]))$, it follows that if $lambda(P) < delta$, then

$$barI leqslant S(f;P) leqslant barI + epsilon$$