If continuous, then Riemann integrable

Theorem 5.4 
If $f:[a,b]\rightarrow R$ is continuous, then $f\in R(x)$ on $[a,b]$.

*The domain of the function is closed and bounded, making it compact.  Having a compact domain implies the function is uniformly continuous, and with this we can say that $f$ is bounded (by previous theorems).


By the definition of uniform continuity, 

          choose an $\epsilon >0$ then there is a $\delta >0$ 

          such that $\left| x-y\right| <\delta$ with $x,y\in [a,b]$ 

          implies $|f(x)-f(y)|<\dfrac {\epsilon}{b-a}$.


Let $P=\{x_0, ..., x_n\}$ be a partition 

          such that $x_i-x_{i-1}<\delta$ for $i=1, ...,n$.

          Then there are $t_i,s_i\in [x_{i-1}, x_i]$ 

          such that $M_i(f)=f(t_i)$ and $m_i(f)=f(s_i)$.

Since $x_i-x_{i-1}<\delta$ then certainly $|t_i-s_i|<\delta$ 

          which implies$|f(t_i)-f(s_i)|<\dfrac {\epsilon}{b-a}$.  

          By a previous theorem, we know $f(t_i)\geq f(s_i)$, so we have a stronger inequality:

          $0\leq f(t_i)-f(s_i)<\dfrac {\epsilon}{b-a}$.

We now have the following:

$U(P,f)-L(P,f)=\sum _{i=1}^{n}[f(t_i)-f(s_i)](x_i-x_{i-1})$

                                $<\sum _{i=1}^{n}\dfrac {\epsilon}{b-a}(x_i-x_{i-1})$

                                $=\dfrac {\epsilon}{b-a}(b-a)=\epsilon$

Therefore, $f\in R(x)$ on $[a,b]$.