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$\alpha$ is monotonic and continuous on [a,b].
Let $\varepsilon$>0 be given then for any positive integer n,
we can choose a partition such that $\Delta\alpha_i=\frac{\alpha(b)-\alpha(a)}{n}$
since $\alpha$ is continuous. (by the theorem)

Theorem: Let f be a continuous real function on the interval [a,b].
If f(a) < f(b) and if c is a number such that f(a) < c < f(b) then there exists a point x belongs to (a,b) such that f(x)=c

I understand $\Delta\alpha_i$ can be chosen like that, but why is this justified by that theorem? I cannot find any relationship between them.

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up vote 1 down vote accepted

I suppose by $\Delta \alpha_i$ you mean: given a partition $a \leq t_0 < \ldots < t_d \leq b$, set $\Delta \alpha_i = \alpha(t_i) - \alpha(t_{i-1})$ for $i =1, \ldots, d$. In order to have $\Delta \alpha_i = \frac{\alpha(b) - \alpha(a)}{n}$, you can choose $t_0 = a$ and then use the (intermediate value) theorem to choose some $t_0 < t_1$ such that $\alpha(t_1) = \alpha(t_0) + \frac{\alpha(b) - \alpha(a)}{n}$ etc.

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