How to use Cauchy criterion to directly prove the integrability of continuous functions?

Question

In Rosenlicht's Introduction to Analysis, he said one can use Cauchy criterion to give a direct proof of the integrability of continuous functions, but I failed to come up with it. Can someone give such a proof?

Answer 1

Suppose $f : [a,b]\rightarrow\mathbb{R}$ is continuous. Then it is uniformly continuous on the closed, bounded $[a,b]$. Let $\epsilon > 0$ be given. Then there exists $\delta > 0$ such that $|f(x)-f(y)| < \epsilon/(b-a)$ whenever $|x-y| < \delta$. Let $\mathcal{P}=\{ p_0 < p_1 < \cdots < p_n \}$ and $\mathcal{Q}=\{ q_0 < q_1 < \cdots < q_m \}$ be partitions of $[a,b]$ for which $\|\mathcal{P}\| < \delta/2$ and $\|\mathcal{Q}\| < \delta/2$. Let $\mathcal{R}$ be the common refinement of $\mathcal{P},\mathcal{Q}$ (without evaluation points.) Then, $$ S(f,\mathcal{P})-S(g,\mathcal{Q}) = \sum_{\mathcal{R}}\{f(p_{n_k}^*)-f(q_{m_k}^*)\}\Delta r_k, $$ where an interval $[r_{k-1},r_{k}]$ lies completely in an interval $[p_{n_{k}-1},p_{n_{k}}]$ and in an interval $[q_{m_{k}-1},q_{m_{k}}]$, with $p_{n_{k}}^*\in [p_{n_k-1},p_{n_k}]$ and $q_{m_{k}}^*\in [q_{m_{k}-1},q_{m_k}]$. Because $p_{n_{k}}-p_{n_{k}-1} < \delta/2$ and $q_{m_{k}}-q_{m_{k}-1} < \delta/2$, then $|p_{n_{k}}^*-q_{m_{k}}^*| < \delta$, which leads to \begin{align} |S(f,\mathcal{P})-S(f,\mathcal{Q})| & \le \sum_{\mathcal{R}}|f(p_{n_k}^*)-f(q_{m_{k}}^*)|\Delta r_k \\ & < \frac{\epsilon}{b-a}\sum_{\mathcal{R}}\Delta r_k=\epsilon. \end{align} Hence, $|S(f,\mathcal{P})-S(f,\mathcal{Q})| < \epsilon$ whenever $\|\mathcal{P}\|, \|\mathcal{Q}\| < \delta$.

Answer 2

Let $\varepsilon > 0$ and $f:[a,b]\to \mathbb{R}$ be a continuous function. The function $f$ also uniformly continuous, hence there exists $\delta > 0$ such that $$|x − y| < δ \implies |f(x) − f(y)| <\frac{\varepsilon}{b − a}$$for all $x, y \in [a, b]$.

Choose a partition of $[a,b]$, say $P = {I_1, I_2, \dots , I_n}$ such that $|I_k| < \delta$ for every $k$. Since $f$ is continuous, it attains its maximum and minimum values, say $f(x_k)=M_k$ and $f(y_k)=m_k$ on the interval $I_k$. And $|I_k|=|x_k-y_k|<\delta$, so
$$M_k − m_k = f(x_k) − f(y_k) <\frac{\varepsilon}{b − a}.$$

The upper and lower sums of $f$ therefore satisfy \begin{align*} U(f; P) − L(f; P) =& \sum_{k=1}^n (M_k − m_k)|I_k|\\ <& \frac{\varepsilon}{b − a} \sum_{k=1}^n |I_k|\\ <& \varepsilon. \end{align*} On the other hand upper and lower integrals $U(f)= \inf_{P\in \Pi} U(f; P)$ and $L(f)= \sup_{P\in \Pi} U(f; P)$ equal ( $\Pi$ denote the collection of all partitions of $[a, b]$) satisfy $$0\leq |U(f)-L(f)|\leq |U(f;P)-L(f;P)|<\varepsilon .$$ Since this inequality holds for all $\varepsilon > 0$, $U(f)=L(f)$ and $f$ is integrable.