# 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 $$f(x) = \begin{cases} 0, & 0 \le x < 1\\ 1, & 1 \le x < 2\\ 2, & 2 \le x < 3\\ \end{cases}$$

1. Prove that $$f$$ is Riemann integrable without appealing to any theorems in this section
2. Which theorems in this section guarantee that $$f$$ is Riemann integrable?
3. 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?

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 $$\lim_{n \to \infty} L_{P(n)}(f) = \int_a^b f(x) dx = \lim_{n \to \infty} U_{P(n)}(f)$$

• But that's using a theorem! Namely the theorem that if $f$ is Riemann integrable on $[a,b]$ and on $[b,c]$ then it is Riemann integrable on $[a,c]$. – David C. Ullrich Nov 1 '18 at 12:56
• @DavidC.Ullrich it only forbids "theorems in this section" -- not sure if this one is included – gt6989b Nov 1 '18 at 16:35
• what if we divide the function into partition P= x0, x1, x2, x3, where x0=0, x1= 1, x2=2 x3=3. Lp(f) on this partition is 4.And Up(f) is 4 as well. Their difference is zero. But the epsilon is greater zero so Up(f)-Lp(f) less than epsilon, which proves the function is riemann integrable. What do you think about this way? – bebe Nov 1 '18 at 18:14
• @bebe it is Riemann integrable when the limit of your mesh becomes the integration interval – gt6989b Nov 1 '18 at 18:16
• @gt6989b Ok then I found a different way. Lets say P = {x0,x1, ..., xN, xN+1,..,xM). and max(xi-xi-1)<ϵ/3. Then Up(f)-Lp(f)<ϵ after some calculation. but I am having problem with finding a theorem to prove that this function f(x) is Riemann integrable. Because I dont know whether it is monotone or not. Is it monotone? – bebe Nov 1 '18 at 18:33

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