How to prove that a Lipschitz function is Riemann integrable on an interval

Question

I am supposed to prove that a Lipschitz function is Riemann integrable. I am thinking that Lipschitz implies continuous, and since it's an interval it's bounded. Is that it?

Answer 1

If $f:[a,b]\to\mathbb R$ is continuous, then it is uniformly continuous. So given $\varepsilon>0$, we may choose $\delta$ such that $|x-y|$ implies $|f(x)-f(y)|<\frac\varepsilon{b-a}$. Let $\mathcal P=\{x_1,\ldots,x_n\}$ be a partition of $[a,b]$ with $$\max_{1\leqslant j\leqslant n-1}(x_{j+1}-x_j)<\delta. $$ Since $[x_j,x_{j+1}]$ is compact, there exist $m_j,M_j\in[x_j,x_{j+1}]$ such that \begin{align} f(m_j) &= \min\{f(x): x\in[x_j,x_{j+1}\}\\ f(M_j) &= \max\{f(x): x\in[x_j,x_{j+1}\}. \end{align} It follows that \begin{align} U_f(\mathcal P)-L_f(\mathcal P) &= \sum_{j=1}^{n-1} [f(M_j)-f(m_j)](x_{j+1}-x_j)\\ &\leqslant \frac\varepsilon{b-a}\sum_{j=1}^{n-1}(x_{j+1}-x_j)\\ &= \frac\varepsilon{b-a}(b-a)\\ &= \varepsilon. \end{align}

Note: If we assumed that $f$ were defined on an open interval $(a,b)$, then Lipschitz continuity yields uniform continuity and so $f$ extends to a continuous function on $[a,b]$ and so there is no loss of generality in assuming that $f$ is defined on a closed interval.