Riemann upper sum of Riemann integral I read this lecture note https://www.math.ucdavis.edu/~hunter/m125b/ch1.pdf. For all partitions $P$ of finite closed interval $[a,b]$, define the upper Riemann sum $f$ w.r.t. partition $P$ by $U(f; P)$. If $f$ is bounded, then $m(b-a)\leq U(f; P)$, where $m\leq f.$

why $\inf_{P} U(f; P)$ exists? We do not need to claim $U(f; P)$ is decreasing?
And how about the $\lim_{P} U(f; P)$? Here is the limit of the net.


The collection of partitions $\mathcal{P}$ is directed set with preorder $\supseteq$ where $P' \supseteq P$ indicates that $P'$ is a refinement of $P$, that is the set of endpoints of subintervals of $P$ is contained in the set of endpoints of subintervals  of $P'$. 
The net $\{U(f;P) \}$ defined on $\mathcal{P}$ is non-increasing in the sense that $P' \supseteq P$ implies that $U(P';f) \leqslant U(P;f)$.  In other words, upper Darboux sums decrease as partitions are refined.
Since the net $\{U(f;P) \}$  is bounded below and non-increasing  it is convergent.
