Rieman integration For this question, I think I need to use the fact that this function is continuous. But, how do I know that? 

Let $f$ be the function on $[0,1]$ given by
  $$
f(x) = \begin{cases}
0, & 0 \le x < 1\\
1, & 1 \le x < 2\\
2, & 2 \le x < 3\\
\end{cases}
$$
  
  
*
  
*Prove that $f$ is Riemann integrable without appealing to any theorems in this section
  
*Which theorems in this section guarantee that $f$ is Riemann integrable?
  
*What is $\int_0^3 f(x) dx$? 
  

Also, how is it possible to prove that this function is Riemann integrable without using any theorems. 
Using that the function is continuous, we could say $U_p(f)-L_p(f) \le \epsilon$ just by saying $|x_n-x_{n-1}|\le \epsilon/3$. And, how do we calculate the integral of it? Is $U_p(f)- L_p(f)$ equal to the integral of f? 
Thank you 
P.S: Every bounded monotone function on a closed interval is Riemann integrable according to the theorem. I cannot find a theorem to show that this function is Riemann integrable because I am not so sure that this function is monotone. Is it monotone increasing?
 A: HINT
It is easier to handle if you say that
$$
\int_0^3 f(x)dx = \int_0^1 f(x)dx + \int_1^2 f(x)dx + \int_2^3 f(x)dx
$$
and then $f$ is continuous on each interval of integration.
As for your second question, I am not sure about your notation, but it seems, given a partition $P(n)$ of $[a,b]$ you have
$$
L_{P(n)}(f) \le \int_a^b f(x) dx \le U_{P(n)}(f)
$$
so if the function is integrable, you end up with
$$
\lim_{n \to \infty} L_{P(n)}(f) = \int_a^b f(x) dx = \lim_{n \to \infty} 
 U_{P(n)}(f)
$$
A: Probably you have to use the definition of Riemann Integral and try to calculate 
$$\sup\{L(f,P):\ P\text{ is a partition of }[0,3]\}$$
and
$$\inf\{U(f,P):\ P\text{ is a partition of }[0,3]\}.$$
I guess that any upper and lower rieman sum will be always the same number, namely, the intergal.
EDIT: at least, you always can add the discontinuity point to your partition to get always the same number. Just an idea.
EDIT2:
Let $P$ be any partition of $[0,3]$. Let $P_0:=P\cup\{1,2\}$. It is clear that $P_0$ is finer than $P$, hence $U(f,P)\ge U(f,P_0)$. But a simple calculation says that $U(f,P_0)=3=\int_0^3 f(x)dx$.
So $$3\le \inf\{U(f,P):\ P\text{ is a partition of }[0,3]\}\le U(f,\{0,1,2,3\})=3.$$
Analogously, you work with $L(f,P)$ and you are done.
It is very easy to get the other inequality
