I need to find a partition. The question is this
by theorem 6.2.2, the function $f(x)=\lfloor{x}\rfloor$ with $x \in [0,4]$ is Riemann integrable. Thus for any $\epsilon >0$, there exists a partition $P$ of $[0,4]$ s.t. $U(p,f)-L(p,f)< \epsilon$. Find such a partition.
Theorem 6.2.2 is "if a function $f:[a,b] \rightarrow \mathbb{R}$ is monotone, $f \in \mathscr{R}[a,b]$
The main things I'm wondering about is if I've computed the upper and lower sums correctly (particularly the upper), and how to continue from here.
I may be doing this wrong, but it seemed like an obvious move to make the partition look like $[0,1] \cup[1,2]\cup[2,3]\cup[3,4]$. So then the lower sum is
$$L(p,f)= \sum m_i (x_i-x_{i-1}) = 0(1-0)+1(2-1)+2(3-2)+3(4-3)$$ which boils down to $0+1+2+3=6$
Since it's a floor function, the upper and lower sums should be the same as $m_i$ and $M_i$ should be the same values. Which also means $U(p,f)-L(p,f)$ should be less than epsilon since for this it would be 0. Is this correct? Any tips for how to continue? Or how to formally write this up?
What you've done so far is fine. Do the same thing for the upper sum. (Or you can simply note that since the function is constant on each sub-interval of the partition, the upper and lower sums will be the same.) Then, as you've noted, any $\varepsilon>0$ will work.