Let $f: [a,b] \rightarrow \mathbb{R}$ be increasing on the set $[a,b]$. Show $f$ is integrable on $[a,b]$

Question

Let $f: [a,b] \rightarrow \mathbb{R}$ be increasing on the set $[a,b]$. Show $f$ is integrable on $[a,b]$.

Note: This has to do with Riemann Integrability

Proof:

Define $M_{k} = f(x_k)$, $m_{k} = f(x_{k-1})$. Also let $U(f,P)$ and $L(f,P)$ to be the upper and lower sum respectively.

Using the $\epsilon$-definition of integrability:

$U(f,P) - L(f,P) = \sum_{k = 1}^n (M_{k} - m_{k-1})\Delta x \\ = \Delta x \sum_{k = 1}^n (f(x_k) - f(x_{k-1})) \\ = \Delta x (b-a)$ (Where does this come from ?)

From here I would arrive at my result by letting $P_{\epsilon}$ beaing a partition with common length satisfying $\Delta x < \frac{\epsilon}{(b-a)}$.

My question is how do we go from $(f(x_k) - f(x_{k-1}))$ to $(b-a)$? I would get it if we were summing the partitions, then I would have a telescoping sum with only the endpoints remaining, but I am summing the values of the function at the specified points, there is nothing that at least to me indicates that the result should be $(b-a)$. Could somebody explain the concept to me?

Answer 1

Observe that you can make: $\triangle x < \dfrac{\epsilon}{f(b) - f(a)}$, and $\displaystyle \sum_{k=1}^n (M_k - m_k) = f(b) - f(a)$. Thus $U(f,P) - L(f,P) < \epsilon\implies f$ is integrable.