Fix $n$ large. Then for $1/n\le x <y\le 1,$ the mean value theorem shows
$$\tag 1 |f(y)-f(x)|\le n^2|y-x|.$$
Now let $m\in \mathbb N$ be greater than $n^3.$ Set
$$P_n=\{0\} \cup\{1/n+k\frac{(1-1/n)}{m},k=0,1,\dots ,m\}.$$
Using $(1),$ we get
$$U(f,P_n)-L(f,P_n)\le \frac{2}{n} +\sum_{k=1}^{m} \left (n^2\cdot \frac{1-1/n}{m}\right)\cdot \frac{1-1/n}{m}$$ $$ \le \frac{2}{n} + m\frac{n^2}{m^2} < \frac{2}{n} + \frac{1}{n} = \frac{3}{n}.$$
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.
Best Answer
The lower and upper sums are sums of areas of rectangles. It should be clear from a diagram that unless $f$ is constant (or at any rate piecewise constant) we will always have $$\{\hbox{lower sum}\}<\{\hbox{exact area}\}<\{\hbox{upper sum}\}\ .$$ For a specific example simply take $f(x)=x$ on $[0,1]$. You should not find it hard to show that for any partition $P$, $$\{\hbox{lower sum}\}<\frac12<\{\hbox{upper sum}\}\ .$$