Answer
$f$ is not integrable function
Work Step by Step
$f$ is not integrable function on a closed interval $[\mathrm{a}, \mathrm{b}]$
\[
f(x)=\left\{\begin{array}{ll}
0 & x \text { is irrational } \\
1 & x \text { is rational }
\end{array}\right.
\]
For every closed interval [a,b], the limit of $\sum_{k=1}^{n} f\left(x_{k}^{*}\right) \Delta x_{k}$ doesn't exist, because if we take $x_{1}^{*}, x_{2}^{*}, \ldots, x_{n}^{*}$ as rational numbers, then we have
\[
\lim _{\max \Delta x_{k} \rightarrow 0} \sum_{k=1}^{n} f\left(x_{k}^{*}\right) \Delta x_{k}=\lim _{\max \Delta x_{k} \rightarrow 0} \sum_{k=1}^{n} \Delta x_{k}=-a+b
\]
If we take $x_{1}^{*}, x_{2}^{*}, \ldots, x_{n}^{*}$ as irrational numbers, then we have:
\[
\lim _{\max \Delta x_{k} \rightarrow 0} \sum_{k=1}^{n} \Delta x_{k} f\left(x_{k}^{*}\right) =0
\]
The limits are not equal.
