Consider a function $f$ and partition $P$ of $[a,b]$. Then the upper sum $U$ of $f$ on $P$ is given by:
\begin{equation} U(f,P)= \sum_{i=1}^n M_i(t_i-t_{i-1}) \end{equation}
Where $M_i=\sup\{ f:t_{i-1}\leq x \leq t_i \}~\forall x\in[a,b]$.
This definition does not bother me. What does, however, is when we consider the following:
\begin{equation} \inf\{{U(f,P)}\} \end{equation}
My understanding of the infimum of a set (this would disturb me for supremum also) is the greatest element that is less than or equal to all the other elements in the set.
But $U(f,P)$ is a summation, one that we would expect to result in a value for a given function $f$ and partition $P$ of $[a,b]$ - a set with one element.
So how is $\inf\{{U(f,P)}\}$ not trivial? (or why isn't it just $U(f,P)$?)
I thought that maybe $\inf\{{U(f,P)}\}$, if it existed, would return the smallest contributing part of the summation, but I haven't seen anything that supports this idea.
I think that in that definition you have to consider the infimum of the set of all the lower sums when $P$ is an arbitrary partition of the interval.