Let $f: [a,b] \rightarrow \mathbb R$ be a function defined on a closed interval $[a,b]$ of the real numbers, $\mathbb R$, and
$\mathcal P = \left \{[x_0,x_1],[x_1,x_2],\dots,[x_{n-1},x_n] \right \}$,
be partition of the closed interval $[a,b]$, where
$ a=x_0<x_1<x_2<\cdots<x_n=b $.
One, perhaps the, way of defining the Riemann upper sum is as follows.
$ U = \displaystyle{\sum_{i=1}^{n} f(x_i^*)\, \Delta x_i }$
where $ \Delta x_i=x_i-x_{i-1} $, $ x_i^*\in[x_{i-1},x_i]$, and $f(x_i^*) = \sup f([x_{i-1},x_i]) $ (that is, the supremum of $f$ over $[x_{i-1},x_i]$).
My question is how does this work for a function with a discontinuity at what would be it's maximum in a particular sub-interval $[x_{i-1},x_i]$; say the function $f$ has a `hole' at the point $c\in [x_{i-1},x_i]$?
For example, take $f(x) = -x^2 + 1$ on $[-1,1]$ with $(0,1)$ removed, and consider a partition with one of the sub-intervals $[-\epsilon, \epsilon]$, $\epsilon >0$. In this situation, what value do you pick for the $x^*$ value in the sub-interval? I mean, it's clear that you'd take a rectangle of height 1 for this bin, since that is the supremum of the set of values of $f$ on this interval, but what value do you take for $x^*$? Does it matter, since the supremum for the set doesn't depend on the chosen value for $x^*$?
I think my misunderstand is how to interpret choosing an $x^*_i \in [x_{i-1},x_i]$ when we are really using the supremum of the set of values of $f$ on $[x_{i-1},x_i]$. Does it matter which value we pick for $x_i^*$, since the supremum of the set is fixed?
Best Answer
It appears that you are having some confusion on Riemann sum, Upper Darboux sum, Lower Darboux sum.
Based on Apostol's Mathematical Analysis, I provide the following definitions which are pretty standard and followed in many other textbooks.
Let $[a, b] $ be a closed interval and let function $f:[a, b] \to\mathbb {R} $ be bounded on $[a, b] $. A partition $P$ of $[a, b] $ is a finite set of points from interval $[a, b] $ and necessarily includes the end points. Typically partition $P$ is written as $$P=\{x_0,x_1,x_2,\dots,x_n\} $$ where $$a=x_0<x_1<x_2<\dots<x_n=b$$ Let us define $$M_k=\sup\, \{f(x) \mid x\in[x_{k-1},x_k]\},k=1,2,\dots,n$$ and $$m_k=\inf\, \{f(x) \mid x\in[x_{k-1},x_k]\}, k=1,2,\dots,n$$ Since $f$ is bounded the numbers $M_k, m_k$ exist. Also since we are not given that $f$ is continuous, these values $M_k, m_k$ may or may not be attained by $f$.
The upper Darboux sum for $f$ over partition $P$ of $[a, b] $, denoted by $U(f, P) $, is defined as $$U(f, P) =\sum_{k=1}^{n}M_k(x_k-x_{k-1})$$ In a similar manner the lower Darboux sum $L(f, P) $ is defined as $$L(f, P) =\sum_{k=1}^{n}m_k(x_k-x_{k-1})$$ Riemann sums are a bit more complicated in the sense that they not only depend upon the partition, but also on a further chosen set of points called tags.
Let $t_1,t_2,\dots,t_n$ be points in $[a, b] $ such that $t_k\in[x_{k-1},x_k]$ for each $k$ and let $$T_P=\{t_1,t_2,\dots,t_n\}$$ The notation $T_P$ is used to emphasize that tag points are chosen based on a given partition.
A Riemann sum for $f$ over partition $P$ of $[a, b] $ with tag points in $T_P$, denoted by $S(f, P, T_P) $, is defined as $$S(f, P, T_P) =\sum_{k=1}^{n}f(t_k)(x_k-x_{k-1})$$
Notice that for any partition $P$ and any tag set $T_P$ we have $$L(f, P) \leq S(f, P, T_P) \leq U(f, P) $$ because $m_k\leq f(t_k) \leq M_k$. Also we can choose the tag points $t_k$ such that $f(t_k) $ is close to $M_k$ (or $m_k$) and hence any upper or lower Darboux can be well approximated by a suitable Riemann sum.
More formally (and try to prove it) if $\epsilon>0$ then we can choose tag sets $T_P, T'_P$ such that $$U(f,P) - \epsilon <S(f, P, T_P) \leq U(f, P), \\ L(f,P) \leq S(f, P, T'_P) <L(f, P)+\epsilon $$ The equality in $\leq $ of these relations given above may or may not occur. In the special case of continuous $f$ such an equality is possible for suitable choice of tags.
Thus you should not try to express an upper Darboux sum as a Riemann sum. They are different but related concepts and it may not be possible to express one as another.
Note: The term Riemann upper sum is not standard and most probably being used in place of the standard term upper Darboux sum.