I've been stuck on this particular problem for a while now:
Let $f: [a,b] \rightarrow \mathbb{R} $ be an increasing function. If $x_1, ... ,x_n \in [a,b]$ are distinct, show that $\sum\limits_{i=1}^n o(f, x_i) < f(b) - f(a)$.
It is clear that this is true, but I am finding it difficult to find a way to prove this in a manner that is valid. The proof that I did have for this problem seems to lack any substance and seems to draw on too many implicit assumptions that makes me almost certain it isn't correct. I think this comes from that fact that I do not see this problem in the way it should be seen, but I am just not exactly sure on what that is. I've tried other approaches that seemed promising at first, though it hasn't been of much help.
I would greatly appreciate any hints or guidance that will point me in the right direction of solving this problem.
Thanks.
Edit:
Note: $o(f, x_i) = \lim_{\delta \to 0} [M(a, f, \delta) - m(a,f, \delta)]$
My proof: Assume $\sum\limits_{i=1}^n o(f, x_i) \geq f(b) - f(a)$. Since $f$ is increasing, we can add $\sum\limits_{i=1}^n o(f, x_i)$ to $f(a)$. But this result is clearly not equal to $f(b)$, because $f$ is defined as increasing, nor is it greater than $f(b)$ because that would be outside of our range. Hence $\sum\limits_{i=1}^n o(f, x_i) < f(b) - f(a)$.
Select points $t_{i-1}<x_i<t_i$ and show that under the assumptions the oscillations satisfy
$$o(f,x_i)\le f(t_i)-f(t_{i-1}).$$
That is, show first the result for $n=1$ and then patch it together for general $n$.