Is the infimum of (Riemann) Upper Sum equal to the limit of Upper Sum as partition size goes to zero? I am following notation in Wheedon and Zygmund, let the (Riemann) upper sum be $U_\Gamma=\sum_{k=1}^N[\sup_{x\in I_k}f(x)]v(I_k)$, and let $\Gamma=\{I_k\}$ a partition of $I$ into finite nonoverlapping intervals, and let $|\Gamma|=\max_k(diam I_k)$.
My question is are the two equal:
1) $\inf_\Gamma U_\Gamma$
2) $\lim_{|\Gamma|\to 0} U_\Gamma$
What I can see is that if $\Gamma'$ is a refinement of $\Gamma$, certainly $U_\Gamma'\leq U_\Gamma$. However it is not immediately clear why the two are equal (or are they)?
Thanks for any help.
 A: Taking $(1)$ as the definition of $\overline \int f$, 
let $\epsilon >0$, and find a partition $\Gamma $ such that
$U_{\Gamma }-\overline \int f <\epsilon$. 
Now take any refinement $\Gamma'$ of $\Gamma.\ $Then, $\vert \Gamma' \vert \le \vert \Gamma \vert $ and
$\overline \int f \le U_{\Gamma' }\le U_{\Gamma }$ which means that 
$U_{\Gamma' }-\overline \int f <\epsilon$.
This is true for $all$ refinements of $\Gamma$, so in fact 
$\lim_{|\Gamma'|\to 0} U_\Gamma'-\overline \int f <\epsilon.$
(One way to see this: choose a sequence of partitions $\Gamma'_n\subseteq \Gamma'_{n+1 }$ such that $\vert \Gamma'_n\vert <\min (1/n,\vert \Gamma \vert) $. Then, $U_{\Gamma_n' }-\overline \int f <\epsilon$ for all $n\in \mathbb N$).
Thus, $(1)$ and $(2)$ are equal.
A: I'll assume $|f| \le M <\infty.$ Denote the infimum over all upper sums by $L.$ Let $\epsilon>0.$ Choose a partition $P$ such that $U(P) < L + \epsilon.$ Let $n$ be the number of subintervals induced by $P.$ We can assume $n>1.$ Finally, let $\delta = \epsilon/[(n-1)(M+1)].$
Now let $Q$ be a partition such that $|Q| < \delta.$ A subinterval of $Q$ may contain a point of $P$ in its interior. There can be no more than $n-1$ such subintervals (thanks to @yoyostein for improving my first estimate of $2n$.) The subsum of $U(Q)$ corresponding to these subintervals is therefore no more than
$$(n-1)M\delta < \epsilon.$$
All the other subintervals of $Q$ are contained in subintervals of $P.$ The subsum of $U(Q)$ corresponding to these subintervals is therefore no more than $U(P).$
Adding these subsums shows
$$L \le U(Q) < \epsilon + (L + \epsilon) = L + 2\epsilon.$$
This gives the desired result.
