Find $$\sup\{f(x)|x\in [\frac{r-1}{n}, \frac{r}{n}]\cap Q^c\}$$ and $$\sup\{f(x)|x\in [\frac{r-1}{n}, \frac{r}{n}]\cap Q\}$$
where $ f(x)=\left\{ \begin{array}{ll} x^2,~~~ x\in Q\cap[0,1]\\ x^3, ~~~x\in Q^c\cap[0,1] \end{array} \right. $ such that $P_n=\{0, 1/n, 2/n, \dots n/n\}$ is a partition of $[0,1]$, $n\in \mathbb{N}$.
I am unable to solve this. I want to find sup in each of the subintervals. Please give me an idea to find the sup.
Personal input: $x^2>x^3$ for $x\in [0,1]$ and hence in each of the subintervals $[\frac{r-1}{n}, \frac{r}{n}]$, $r=1,2,3,...,n$. So $$\sup\{f(x)|x\in [\frac{r-1}{n}, \frac{r}{n}]\cap Q\}$$ would be $\frac{r^2}{n^2}$ as $f(x)=x^2$ is monotone increasing on $[\frac{r-1}{n}, \frac{r}{n}]\cap Q$.
Correct? What about other one?
So $x\rightarrow x^{3}$ is increasing on $[0,1]$, so we target the $(r/n)^{3}$ if possible. For each $x\in[(r-1)/n,r/n]\cap{\bf{Q}}^{c}$, we have $f(x)=x^{3}\leq(r/n)^{3}$, so $\sup\{\cdots{\bf{Q}}^{c}\}\leq(r/n)^{3}$. And we can choose $(p_{n})\subseteq{\bf{Q}}^{c}$ such that $p_{n}\in[(r-1)/n,r/n]$ and that $p_{n}\rightarrow r/n$. For a small $\epsilon>0$, choose an $n$ such that $(r/n)^{3}-\epsilon<p_{n}^{3}$, this shows that $(r/n)^{3}\leq\sup\{\cdots{\bf{Q}}^{c}\}$ and hence $\sup\{\cdots{\bf{Q}}^{c}\}=(r/n)^{3}$.