Let $f: [-2,2] \to \mathbb R$, $$ f(t) = \begin{cases} -1 \, &t<0 \\ 1 \, &t\ge 0 \end{cases} $$
Define $g: [-2,2] \to \mathbb R$ as: $$g(x) = \int_{-2}^xf(t)dt$$
Plot $g(x)$ and find it's limit at 0.
Based on my reading from Wiki, I need to check whether the function is Riemann integrable (which it should be based on graphical intuition). The doubt is that there is only one sub-interval, of some given partition, due to which the lower and upper Darboux sums will not match.
But how do I show in the limit they are same?
EDIT: Based on learnings from various comments, another attempt:
For a given $x \ge 0$, consider the partition of $[-2,x]$ in $2n$ parts such that all the sub-intervals in $[-2,0]$ are of equal length $2/n$ and the rest of the partitions are of length $x/n$. Call this partition $P$.
\begin{align} U(P,f,x) &= \lim_{n \to \infty} \bigg[(n-1)\frac{2}{n}(-1) + \frac{2}{n}(1) + (n-1)\frac{x}{n}(1)\bigg] \\ &=x-2 \\ L(P,f,x) &= \lim_{n \to \infty} \bigg[(n\frac{2}{n}(-1) + \frac{2}{n}(1) + (n-1)\frac{x}{n}(1)\bigg] \\ &= x-2 \end{align}
For $x<0$, $f(x)$ is continuous so $g(x)$ is well defined and is $-2-x$.
As shown above, since in the limit Darboux sums are same, $g(x)$ is defined and is equal to the limit of the Darboux sum. Which gives:
$$g(x) = |x|-2$$
Is this the right approach?
EDIT 2: In terms of $\epsilon$:
Let $P$ be defined in the same way except that $n>4/\epsilon$. Now, for $P$,
$$U(P,f) - L(P,f) = 4/n < \epsilon$$
So for any $\epsilon >0$, $\exists \,\, P$, such that $U=L$