One refinement worked out for $f(x) = x^{2}$ on $[0,1]$: adding the point $1/2$ to the trivial partition raises the lower sum from $0$ to $1/8$ and lowers the upper sum from $1$ to $5/8$

Let $f : [0,1] \to \mathbb{R}$ be $f(x) = x^{2}$ (Integer powers $a^m$). Let $P_0$ be the trivial partition of $[0,1]$, with point set $\{0,1\}$, and let $P_1 := P_0 + \tfrac12$ be the partition obtained by inserting the point $1/2$, with point set $\{0, \tfrac12, 1\}$ (Partition of $[a,b]$ as a finite strictly increasing list $a = t_0 < t_1 < \dots < t_n = b$, its subintervals and their lengths, its mesh, refinement, and the common refinement of two partitions). Then

$L(f,P_0) = 0, \qquad L(f,P_1) = \tfrac18, \qquad U(f,P_1) = \tfrac58, \qquad U(f,P_0) = 1 ,$

so that

$0 \;=\; L(f,P_0) \;\le\; L(f,P_1) \;\le\; U(f,P_1) \;\le\; U(f,P_0) \;=\; 1 ,$

which is claim 1 of Refining a partition raises the lower Darboux sum and lowers the upper one, and every lower sum is at most every upper sum: $L(f,P) \le L(f,P') \le U(f,P') \le U(f,P)$ when $P'$ refines $P$, and $L(f,P) \le U(f,Q)$ for arbitrary partitions $P$ and $Q$; moreover the two changes are at most $2M(n' - n)\|P\|$ with both inequalities strict. The gap drops from $U(f,P_0) - L(f,P_0) = 1$ to $U(f,P_1) - L(f,P_1) = \tfrac12$: exactly one refinement halves it, and $\int_0^1 x^{2} = 1/3$, computed from the Darboux definition with uniform partitions and the closed form $\sum_{k<n} k^{2} = n(n-1)(2n-1)/6$ shows the uniform partitions drive it to $0$.

• Refinement is an improvement, never a deterioration. That is the content of Refining a partition raises the lower Darboux sum and lowers the upper one, and every lower sum is at most every upper sum: $L(f,P) \le L(f,P') \le U(f,P') \le U(f,P)$ when $P'$ refines $P$, and $L(f,P) \le U(f,Q)$ for arbitrary partitions $P$ and $Q$; moreover the two changes are at most $2M(n' - n)\|P\|$ and it is what makes The lower and upper Darboux integrals of a bounded $f$ on $[a,b]$ as $\sup_P L(f,P)$ and $\inf_P U(f,P)$, Darboux integrability as their equality, and the notation $\int_a^b f$ well posed: the lower sums increase and the upper sums decrease, so the supremum of the one and the infimum of the other are the right things to take.

• A single insertion cannot close the gap. Here it halves it, from $1$ to $\tfrac12$, and no finite number of insertions makes $U - L$ zero for a non-constant $f$: each $M_i - m_i$ is positive whenever $f$ is non-constant on $I_i$. Integrability is the statement that the gap can be made arbitrarily small, not zero, which is exactly the form of Riemann's criterion: a bounded $f$ on $[a,b]$ is Darboux integrable if and only if for every real $\varepsilon > 0$ there is a partition $P$ with $U(f,P) - L(f,P) < \varepsilon$.

• The values are exact rationals, and worth checking by hand. The lower sum of the refined partition is $\tfrac18 = 0.125$ and the upper is $\tfrac58 = 0.625$; the true integral is $\tfrac13$ ($\int_0^1 x^{2} = 1/3$, computed from the Darboux definition with uniform partitions and the closed form $\sum_{k<n} k^{2} = n(n-1)(2n-1)/6$), which indeed lies strictly between them.