Given $$g(x)=\begin{cases} -2 &-1\leq x\lt 0\\ 1 & x=0 \\ 2 & 0 \lt x \leq 3 \end{cases} $$ I need to prove $g$ is Darboux integrable on $[-1,3]$. When I read Bartle's book, I feel like that I need to define partition and my idea is $\mathcal{P}=(-1,-\epsilon,\epsilon,3)$ for $\epsilon \gt0$ then try to find the lower sum and upper sum from the partition. When I tried to compute, I found this
$$L(f;\mathcal{P})=-2(-\epsilon-(-1))-2(\epsilon - (-\epsilon)+2(3-\epsilon) = 4 - 4\epsilon \\ U(f;\mathcal{P})=-2(-\epsilon-(-1))+2(\epsilon - (-\epsilon)+2(3-\epsilon)=4+4\epsilon$$
Is this approach correct? If not, what is the correct approach? If this approach is correct :
- Even though I came up with such partition, I still confused. Is such partition correct? Should we include $x=0$ inside the partition? Generally, how do we come up with such partition?
- After finding the lower sum and upper sum, what I saw in Bartle's book is saying that $U(f) \leq something$, and $L(f) \geq something$. How do we conclude this?
I'm sorry if I asked too many questions. It's okay if you can't answer all of it. Thank you in advance!