Given $$g(x) = |3x-1|$$ For each $ n \in \mathbb{N}$, let $P_n = \{-1, -1 + \frac{1}{3n}, -1, \frac{2}{3n}, ..., 1\}$ be an evenly spaced partition of [-1,1] so that each interval in the partition has length $\frac{1}{3n}$.
This question require to find the Darboux sums $U(g, P_n) and L(g, P_n)$ for each $n \in \mathbb{N}$.
and hence determine if g is integrable.
I have no idea how to handle this type of question. Do I need to consider $ n = -1, 0$ and $ 1 $ only?
a remarks, can I show the integrability by stating g is monotone? But this question needs to use the Darboux sums as well..
We can find an upper function $\phi_+$ as $$\phi_+=\begin{cases} g(x_k)&\text{if $x\in(x_k,x_{k+1})$, $k=0,1,...,4n-1$}\\ g(x_{k+1})&\text{if $x\in(x_k,x_{k+1}),k=4n,4n+1,...,6n-1$} \end{cases}$$ where $x_k$ is the $k^{th}$ element in the partition $P_n$. A similar lower function can be found, and we can easily compute $$I(\phi_+)-I(\phi_-)= \sum_{k=0}^{4n-1}(g(x_k)-g(x_{k+1}))\frac{1}{3n} -\sum_{k=4n}^{6n}(g(x_{k+1})-g(x_{k}))\frac{1}{3n} =\sum_{k=0}^{4n-1}(4-\frac{k}{n}-(4-\frac{k+1}{n}))\frac{1}{3n} -\sum_{k=4n}^{6n}(\frac{k+1}{n}-4-(\frac{k}{n}-4))\frac{1}{3n} =\sum_{k=0}^{4n-1}\frac{1}{3n^2} -\sum_{k=4n}^{6n}\frac{1}{3n^2}$$$$ =\frac{4n-1}{3n^2}-\frac{2n}{n^2}$$ and this right hand expression tends to zero as $n\to \infty$ so $g$ is integrable.
To clarify:
$\phi_+$ is an upper function for $g$ because $\phi_+\geq g,\forall x\in[-1,1]$. Similar for $\phi_-$.
Noting that the function is decreasing on $[-1,\frac{1}{3}]$, the left hand side of each interval in the partition will have the maximum value so we give it to $\phi_+$ here, and we give the minimum value (right hand side) to $\phi_-$. At $x=\frac{1}{3}$, $g$ is zero and then becomes increasing on $[\frac{1}{3},1]$ so we assign values the opposite way here.
Note also we can write $x_k=-1+k(\frac{1}{3n})$ for each member of $P_n$ - there are $6n+1$ in total and $k=4n$ corresponds to the turning point $x=\frac{1}{3}.$
$I$ is just the function for computing the area underneath a step function (what you call the $U$ or $L$ functions) so we have shown that the difference in area between a function greater than $g$ and another function less than $g$tends to zero as the functions get closer and closer to $g$ or equivalently $n\to \infty$.