It is well-known that in case of a linear parititon of $[0,1]$, $\{t_n\}_{n=1}^N = \{\frac{n}{N}\}_{n=1}^N$, we have $$ \int_{t_n}^{t_{n+1}} dt = t_{n+1} - t_n = \frac{1}{N} \quad \forall n$$
But there exists a trajectory $g(t)$ such that $$ \int_{t_n}^{t_{n+1}} dg(t) = \frac{1}{N^\alpha} \quad \forall n$$ for $\alpha \in (0,1]$, given the same uniform partition on $[0,1]$? Is there a well-known class of functions satisfying this condition?
I found a negative solution to the question.
Suppose such a function exists, then $$F(t_{n+1}) - F(t_n) = \int_{t_n}^{t_{n+1}} dg(t) = \frac{1}{N^\alpha}$$ assuming $F(t_0) = 0$, we can construct the solution on the point of the partition, i.e. $F(t_1) = \frac{1}{N^\alpha}, F(t_2) = \frac{2}{N^\alpha}$ and so on, until we get $F(t_N) = N^{1-\alpha}$. However, if $N = 1$, then $F(t_1) = F(t_N) = 1$. Because such function has to be unique over the choice of $N$, which represents the size of the partition on $[0,1]$, we can choose $N$ as large as we want, we send it to infinity, hence $$1 = F(1) = \lim_{N\to\infty} F(1) = \lim_{N\to\infty} N^{1-\alpha} = \infty$$ achieving a contradiction. Therefore proving that such a function cannot exist as $\alpha \neq 1$.