How to show $f$ is Riemann Integrable, finding the $m_i$ and $M_i$ values?

Question

I keep getting stuck on the same sort of question on Riemann Integrals, I am trying to show that a function f is Riemann Integrable on an interval. e.g. Let $f : [−4, 4] \to \mathbb{R} $ be the function given by:

• $2$ if $-4 \leq x <1$

• $1$ if $x=1$

• $0$ if $1<x \leq 4$

Using the partition, $P_n$=$ (-4,1-2/n,1,1+2/n,4) $

Calculate $L(P_n)$ and the upper sum $U(P_n)$. Hence show that $f$ is Riemann Integrable.

What values of $m_i$ and $M_i$ should I use and how do I work these values out? Thanks in advance, I understand how to formulate the answer just not sure how to find these values to do it.

Answer 1

The partition is given, so you have the partition (sub)intervals. Use left ends of the subintervals for your $m_i$ and the right ends for your $M_i$. The function is a piecewise constant, so you should not have too much trouble to find the sums explicitly and then the $\lim_{n\to\infty}$ of the sums which will be the same. Interestingly, the value of the function at 1 is irrelevant.

Answer 2

For a arbitrary $\epsilon > 0$:

$L(P_n) = 2\left(1-\dfrac{2}{n} - (-4)\right)+ 1\left(1-\left(1-\dfrac{2}{n}\right)\right)+ 0\left(\left(1+\dfrac{2}{n}\right) - 1\right)+ 0\left(4-\left(1+\dfrac{2}{n}\right)\right)=10-\dfrac{2}{n}$

$U(P_n) = 2\left(1-\dfrac{2}{n}-(-4)\right)+2\left(1-\left(1-\dfrac{2}{n}\right)\right)+1\left(\left(1+\dfrac{2}{n}\right)-1\right)+0\left(4-\left(1+\dfrac{2}{n}\right)\right)$.

Thus: $U(P_n) - L(P_n) = \dfrac{4}{n}< \epsilon\iff n > \dfrac{4}{\epsilon}$.

This means $f$ is integrable.