Any continuous function is Riemann-integrable on a closed interval

The goal of this post is to prove one of the practical foundations of Riemann-integrals.

Theorem

A function that is continuous on $[a,b]$ is integrable on $[a,b]$.  

See also: Riemann integral

See also: Uniformly continuous functions

Proof. Let $f(x)$ be continuous on $[a,b]$. By the above theorem, $f(x)$ is uniformly continuous on $[a,b]$. Therefore, for any $\epsilon >0$, there exists a $\delta >0$ such that, for all $x,y\in [a,b]$, $|x-y|<\delta$ implies $|f(x)-f(y)|<\frac{\epsilon }{b-a}$.

Let $\mathrm{\Delta }$ be a partition of $[a,b]$ such that $a=x_0<x_1<\cdots <x_{n-1}<x_n=b$ and its mesh is less than $\delta$ (i.e., $x_{i+1}-x_i<\delta$ for all $i=0,1,\cdots ,n-1$). Then, for each $i=0,1,\cdots ,n-1$, if $x,y\in [x_i,x_{i+1}]$, then $|f(x)-f(y)|<\frac{\epsilon }{b-a}$. Hence, if we define

$$\begin{array}{rcl}{M}_i& =& sup\{f(x)\mid x_i\le x\le x_{i+1}\},\\ m_i& =& inf\{f(x)\mid x_i\le x\le x_{i+1}\},\end{array}$$

$M_i-m_i<\frac{\epsilon }{b-a}$. Therefore, using the upper and lower Riemann sums,

$$\begin{array}{rcl}{S}_{\mathrm{\Delta }}& =& \sum _{i=0}^{n-1}{M}_i(x_{i+1}-x_i),\\ s_{\mathrm{\Delta }}& =& \sum _{i=0}^{n-1}{m}_i(x_{i+1}-x_i),\end{array}$$

we have

$$\begin{array}{rcl}{S}_{\mathrm{\Delta }}-s_{\mathrm{\Delta }}& =& \sum _{i=0}^{n-1}(M_i-m_i)(x_{i+1}-x_i)\\ & <& \sum _{i=0}^{n-1}\frac{\epsilon }{b-a}(x_{i+1}-x_i)\\ & =& \frac{\epsilon }{b-a}\cdot (b-a)=\epsilon .\end{array}$$

Thus, in particular,

$$\underset{\mathrm{\Delta }}{inf}{S}_{\mathrm{\Delta }}-\underset{\mathrm{\Delta }}{sup}{s}_{\mathrm{\Delta }}<\epsilon .$$

However, since $\epsilon >0$ is arbitrary, it follows that

$$\underset{\mathrm{\Delta }}{inf}{S}_{\mathrm{\Delta }}=\underset{\mathrm{\Delta }}{sup}{s}_{\mathrm{\Delta }}.$$

Therefore, $f(x)$ is Riemann-integrable. ■