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

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!

• Use $\delta > 0$ as an upper bound for the norm of the partition $P$, and sum on $M_i - m_i$, not on $x_i - x_{i-1}$. Commented Dec 7, 2015 at 5:55
• @GNUSupporter Do you mean this? $$\delta \sum_{i=1}^n(M_i−m_i)$$
– mxdg
Commented Dec 7, 2015 at 6:04
• Bravo ! You've got it! Commented Dec 7, 2015 at 6:04
• @GNUSupporter So, to complete the proof, do I then say that I can choose $\delta = \epsilon/\sum_{i=1}^n(M_i - m_i)$? But this $\delta$ is set by the definition of uniformly continuous (which is how we get the upper bound), so I can't reset it, right?
– mxdg
Commented Dec 7, 2015 at 6:11
• I'm sorry for remembering the proof in a wrong way. I've just read the proof from a book. Actually, we should set $M_i$ and $m_i$ to be something like $f(v_i)$ and $f(u_i)$ in the domain, and bounded it by (by Extreme value Theorem) of $f(v) - f(u)$, where $u$ and $v$ are the minimizer and maximizer of $f$. Then we sum on $x_i - x_{i-1}$ Commented Dec 7, 2015 at 6:16

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 $S^*(f,P) - S_*(f, P) < \epsilon$.

• In the third to last line, did you mean to put $\leq \sum_{i=1}^n K\delta(x_i-x_{i-1})$, rather than $\leq \sum_{i=1}^n K(x_i-x_{i-1})(x_i-x_{i-1})$ ?
– mxdg
Commented Dec 8, 2015 at 0:14
• I think both are OK because $u_i, v_i$ are points in the partition $[x_{i-1},x_i]$, and the length of this partition $x_i - x_{i-1} \le \delta$. Commented Dec 8, 2015 at 2:03