Show that the sum of the oscilations is less or equal to $f(b)-f(a)$

Question

I want to show the following:

Let $ f:[a,b]\to \mathbb{R}$ be an increasing function.If $ x_1,\ldots,x_k\in[a,b]$ are different, show that $$\displaystyle\sum_{i=1}^k o(f,x_i) < f(b) - f(a).$$

My attempt:

Since $f$ is increasing we have that:

$$f(a)<f(x_1)<f(x_2)< \ldots <f(b).$$

Therefore for each $i=1, \ldots, k$ and for all $\delta$,

$$M(f,x_i, \delta)< f(b) \quad \text{and} \quad m(f,x_i, \delta)> f(a)$$

and then in each point:

$$o(f,x_i) < f(b) - f(a).$$

The thing is that I don't know how to proceed with the sum. Can someone help me to complete this proof?

Answer 1

We should assume that $f$ is not constant at $[a,b]-\{x_1,\ldots,x_n\}$.

Choose $a=y_0\le x_1<y_1<x_2<\cdots<x_n\le y_n=b$, then $O(f,x_i)\le f(y_i)-f(y_{i-1})$. Add these inequalities together to get what you want.