proof of continuous functions are Riemann integrable

Recall the definition of Riemann integral. To prove that $f$ is integrable we have to prove that $\\lim_{\\delta\\to 0^{+}}S^{*}(\\delta)-S_{*}(\\delta)=0$ .Since $S^{*}(\\delta)$ is decreasing and $S_{*}(\\delta)$ is increasing it is enough to show that given $\\epsilon>0$ there exists $\\delta>0$ such that $S^{*}(\\delta)-S_{*}(\\delta)<\\epsilon$ .

So let $\\epsilon>0$ be fixed.

By Heine-Cantor Theorem $f$ is uniformly continuous i.e.

\\exists\\delta>0\\ |x-y|<\\delta\\Rightarrow|f(x)-f(y)|<\\frac{\\epsilon}{b-a}.

Let now $P$ be any partition of $[a,b]$ in $C(\\delta)$ i.e. a partition $\\{x_{0}=a,x_{1},\\ldots,x_{N}=b\\}$ such that $x_{i+1}-x_{i}<\\delta$ . In any small interval $[x_{i},x_{i+1}]$ the function $f$ (being continuous) has a maximum $M_{i}$ and minimum $m_{i}$ . Since $f$ is uniformly continuous and $x_{i+1}-x_{i}<\\delta$ we have $M_{i}-m_{i}<\\epsilon/(b-a)$ . So the difference between upper and lower Riemann sums is

\\sum_{i}M_{i}(x_{i+1}-x_{i})-\\sum_{i}m_{i}(x_{i+1}-x_{i})\\leq\\frac{\\epsilon}{b%-a}\\sum_{i}(x_{i+1}-x_{i})=\\epsilon.

This being true for every partition $P$ in $C(\\delta)$ we conclude that $S^{*}(\\delta)-S_{*}(\\delta)<\\epsilon$ .