I have the following homework problem:
Find the upper sum and lower sum for the following function with respect to the given partition:
Let $s:[0,1]\rightarrow \Bbb R$ be defined by: $$s(x)=\bigg\{^{\frac{1}{q} \space if\space x\in \Bbb Q \cap \big[0,1\big] and\space x=\frac{p}{q} \space in\space lowest\space terms,\space where\space p\in \Bbb N \cup \{0\}\space and\space q\in \Bbb N}_{0, \space otherwise}$$
Let $Q=\{ 0,\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5},1\}$
So, if I understand this correctly:
x is restricted to the rationals between zero and one, inclusive, so $x\in \Bbb Q:0\leq x\leq 1$
p is the natural numbers with zero included, which is the same as a non-negative integer, so it could be rewritten $p\in \Bbb Z:p\geq0$, or more simply $p\in \Bbb Z^+$
So, $x=\big\{\frac{p}{q}:p\in \Bbb Z^+,q\in \Bbb N\big \}$.
Attempt #1:
$U(f,Q)=\sum^{n}_{i=1}M_i(f)(x_i-x_{i-1})$
$L(f,Q)=\sum^{n}_{i=1}m_i(f)(x_i-x_{i-1})$
$(x_i-x_{i-1})=\frac{1}{5}$
$U(f,Q)=(\frac{1}{5}\times \frac{1}{5})+(\frac{2}{5}\times \frac{1}{5})+(\frac{3}{5}\times \frac{1}{5})+(\frac{4}{5}\times \frac{1}{5})+(1\times \frac{1}{5})=\frac{3}{5}$
$L(f,Q)=(0\times \frac{1}{5})+(\frac{1}{5}\times \frac{1}{5})+(\frac{2}{5}\times \frac{1}{5})+(\frac{3}{5}\times \frac{1}{5})+(\frac{4}{5}\times \frac{1}{5})=\frac{2}{5}$