TITLE: Does $\lim \limits_{n \to \infty} U(f,P_n) = \inf_{p \in \mathcal{P}}\ U(f,p)$?
QUESTION [2 upvotes]: I've seen this written several different ways in class and in Bruckner's Elemenatry Real Analysis, but I am not convinced they are equivalent. 
Does $\lim \limits_{n \to \infty} U(f,P_n) = \inf_{p \in \mathcal{P}}\ U(f,p)$?
Let $f : [a,b] \to \mathbb{R}$.
Where $\mathcal{P}$ is the set of all partitions $p$ of $[a, b] $. $U(f,P)$ is the upper sum, i.e. $U(f,P) = \sum_{k=1}^n \sup\{f(x)\ : x \in [x_{k-1},x_k]\}(x_k-x_{k-1})$.
$(P_n)$ is defined as a sequence of partitions of $[a,b]$ such that $\|P_n\| \to 0$

REPLY [2 votes]: Assuming only that $f$ is bounded, let $|f(x)| \leqslant M$ for all $x \in [a,b]$ and let $A = \inf_P U(f,P)$.
By definition of the infimum, for any $\epsilon > 0$ there is a partition $P'$ such that
$$A \leqslant U(f,P') < A + \epsilon/2$$
Let $δ=ϵ/4mM\,$  where $m$  is the number of points in the partition $P'$, and  let $P$  be any partition with $||P|| < \delta$ . Form the common refinement $Q=P∪P'$  .
You will see that the upper sums $U(f,P)$  and $U(f,Q)$  differ in at most $m$  subintervals and in each the deviation is bounded by $2M\delta$.
Thus, 
$$U(f,P) - U(f,Q) \leqslant|U(P,f)-U(Q,f)| < m2M\delta =m2M\frac{\epsilon}{4mM}=\epsilon/2,$$
and, since $Q$ is a refinement of $P'$ implying $U(f,Q) \leqslant U(f,P')$, we have 
$$A \leqslant U(f,P) < U(f,Q) + \epsilon/2 \leqslant U(f,P') + \epsilon/2 < A + \epsilon.$$
Choosing $N$ such that $\|P_n\| < \delta $ for all $n \geqslant N$, we have 
$$A \leqslant U(f,P_n) < A + \epsilon$$
Therefore, 
$$\lim_{n \to \infty} U(f,P_n) = A = \inf_P U(f,P).$$
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