Prove Lipschitz function $f$ with constant $K$ is integrable on $[0, 1].$

Question

We suppose that $f : [0, 1] \rightarrow \mathbb{R}$ is a Lipschitz function with constant $K$. We want to show that $f$ is integrable on $[0, 1].$

I've been trying to use the Darboux criterion of integrability by trying to show that, for $\epsilon > 0$, $$S^*(f,P) - S_*(f, P) < \epsilon,$$ where $S^*(f, P)$ and $S_*(f, P)$ are the Upper and Lower Riemann Sums, respectively.

This is what I have so far: $$\sum_{i=1}^n M_i (x_i - x_{i-1})- \sum_{i=1}^n m_i (x_i - x_{i-1}) = \sum_{i=1}^n (M_i - m_i) (x_i - x_{i-1})$$ $$\le \sum_{i=1}^n K (x_i - x_{i-1}).$$ because $|f(x) - f(u)| \le K|x-u| \le K |1|$, since $x\in [0, 1]$. Then, $$\sum_{i=1}^n K (x_i - x_{i-1}) = K.$$ However, this is only a single upper bound, not an arbitrary $\epsilon.$ Can I get some pointers about how to proceed? Thanks!

Answer 1

Since I see the message "Please avoid extended discussions in comments ...", I decided to post my response here. Let $\epsilon > 0$. Suppose $\delta > 0$ is the norm of the partition $P$.

By Extreme Value Theorem, for each partition, we can choose $u_i,v_i \in [x_{i-1},x_i]$ such that $m_i = f(u_i) \le f(x) \le f(v_i) = M_i \forall x \in [x_{i-1},x_i]$.

\begin{align*} &\sum_{i=1}^n M_i (x_i - x_{i-1})- \sum_{i=1}^n m_i (x_i - x_{i-1})\\ =& \sum_{i=1}^n (M_i - m_i) (x_i - x_{i-1}) \\ =& \sum_{i=1}^n (f(v_i)-f(u_i)) (x_i - x_{i-1}) \\ \le& \sum_{i=1}^n K|v_i-u_i| (x_i - x_{i-1}) \\ \le& \sum_{i=1}^n K(x_i - x_{i-1}) (x_i - x_{i-1}) \\ \le& K\delta \sum_{i=1}^n (x_i - x_{i-1}) \\ =& K\delta \end{align*}

So set $\delta = \dfrac\epsilon{K}$. Then $S^*(f,P) - S_*(f, P) < \epsilon$.