$$g(x)= \begin{cases} 6 &\text{, if x $\in \mathbb{Q}$}\\ x &\text{, if x $\notin$ $\mathbb{Q}$} \end{cases}$$
If $P$ is a partition: $\{x_0 = 7.1, x_1, x_2, \dots , x_n = 8\}$, find $L(g,P)$ and $U(g,P)$. Simplify sums where possible.
$m_i$ = inf $\{g(x), \text{ all partitions P on [7.1, 8]}\}$
We know that $g$ will take on the value $6$. But we have that $g(x)$ takes on $'x'$ as well, and I'm not sure how to handle this?
Consider a general partition of $[7.1,8]$ that is not necessarily uniform. Since the rationals are dense in any interval, it follows that on each subinterval $[x_{j-1},x_j]$
$$\sup_{x \in [x_{j-1},x_j]}g(x) = \max(x_j,6) = x_j, \\ \inf_{x \in [x_{j-1},x_j]}g(x) = \min(x_{j-1},6) = 6.$$
The upper sum is
$$\begin{align}U(g,P) &= \sum_{j=1}^nx_j(x_j - x_{j-1}) \\ &= \frac{1}{2}\sum_{j=1}^n(x_j+ x_{j-1})(x_j - x_{j-1}) + \frac{1}{2}\sum_{j=1}^n(x_j - x_{j-1})(x_j - x_{j-1}) \\ &= \frac{1}{2}\sum_{j=1}^n(x_j^2 - x_{j-1}^2)+ \frac{1}{2}\sum_{j=1}^n(x_j - x_{j-1})^2 \\ &= \frac{x_n^2 - x_0^2}{2 } + \frac{1}{2}\sum_{j=1}^n(x_j - x_{j-1})^2 .\end{align}$$
The second term cannot be further reduced without providing more details on the partition.
Finding the lower sum is straightforward.