# Fitzpatrick's proof of Darboux sum comparison lemma

I am just reading Fitzpatrick's advanced calculus. He wants to prove $\lim (\max(x_{i-1} - x_i)) =0$ and $\lim(U(f,P)-L(f,P))$ is equivalent to $f$ is integrable.

He used darboux sum comparison lemma, which states for a bounded function $f$, for a given partition P with k partitions points, the following inequality holds for any partition $P^{*}$ of $[a, b]$:

$U(f,P^{*}) \leqslant U(f,P) + k M gap(P^{*})$, where $gap(P^{*})$ is $\max(x_{i-1} - x_i)$, $M$ is the bound of $f$.

You can refer to lemma 7.10 at google book http://goo.gl/MHs8pB.

What I don't understand in the proof is, he has written $\sum _{i\notin C} M_i (x_i - x_{i-1}) \leqslant U(f,P^{'})$, where $P^{'}$ is a refinement of $P$ and $P^{*}$.

I know that $x_i$ and $x_{i-1}$ for $i\notin C$ are partition points of $P^{'}$. However, $U(f,P^{'}) = \sum M_i (x_i - x_{i-1}) = \sum _{i\notin C} M_i (x_i - x_{i-1}) + \sum _{i \in D} M_i (x_i - x_{i-1})$, where D contains other partition points of $P^{'}$ excluding endpoints.

And my worry is if some $M_i$ are negative, then the inequality $\sum _{i\notin C} M_i (x_i - x_{i-1}) \leqslant U(f,P^{'})$ may not hold.

Please help.

## 1 Answer

Denote $C$ as the set of crossing indices, as defined earlier in Patrick's proof.

It follows that,$\ \sum_{i \in C}M_i(x_i-x_{i-1}) \leq kM\text{gap}P^{*}$ and $\sum_{i \in C}m_i(x_i-x_{i-1}) \geq -kM\text{gap}P^{*} \ .$

On other hand, if $i$ is not crossing index, then $[x_{i-1},x_i] \subset [z_{j-1},z_j],$ for some $j.$ $\implies \sum_{i \not\in C}(M_i-m_i)(x_i-x_{i-1})\leq \sum_{j: \ {[x_{i-1},x_i] \subset [z_{j-1},z_j]}}(K_j -k_j)(z_{j}-z_{j-1})$ $\leq U(f,P) -L(f,P),$ $\text{where} \ K_j (\text{resp.} \ k_j) \ \text{is sup (resp. inf) of} \ f \ \text{in} \ [z_{j-1},z_j].$

Hence, $U(f,P^*)-L(f,P^*) = \sum_{i \in C}(M_i-m_i)(x_{i}-x_{i-1})+\sum_{i \not\in C}(M_i-m_i)(x_{i}-x_{i-1})$ $\leq 2E+ U(f,P)-L(f,P).$