How to prove that a function is integratable using the Riemann condition?

Question

Specifically, From reading a some texts, it states that the Riemann condition for showing that the function is integratable is by showing that there is an $\epsilon > 0$ and condition $s \leq f \leq t$ (s and t are step functions below and above f respectively). To prove integrability it is said to then show that $$ \int_{a}^{b}t-\int_{a}^{b}s < \epsilon $$ However, the steps to get this point are somewhat unclear and I am not sure where to start. For example, how do I show that $x^2$ is integratable from $[0, 2]$ using this condition? Is it perhaps possible to show integrability through the $\epsilon-\delta$ definition of continuity? If so, how do I go about doing that? Is there a good intuition to help understand this method? Any help is appreciated.

Answer 1

The intuition is the same as for proving continuity with $\varepsilon-\delta$, i.e. you start of with the scratch work and when you reach the end, you know what you require $\delta$ to be. In the case of integrals, after expanding the difference of the upper sum and lower sum for some partition $P$, you will know how fine the partition needs to be, so that for any given $\varepsilon>0$, you have $U(f,P)-L(f,P)<\varepsilon$ whenever $\lvert P\rvert <\delta$.

For your case of $f(x)=x^2$, note that $f$ is increasing on $[0,2]$, so the supremum of $f$ for any sub-interval is the right endpoint, and the infimum is the left endpoint. When evaluating the difference we get a telescoping sum, which leaves the right-most end point and left-most endpoint, times the length of the subintervals. The proof proceeds as follows:

Let $\varepsilon>0$ be given and let $P$ be a partition of $[0,2]$ into $n$ parts as follows: $0=x_0<x_1<\ldots<x_{n-1}<x_n=2$, where $n$ is the smallest natural number for which $n>8/\varepsilon$. Thus we get $$ U(f,P)-L(f,P)=\sum_{k=1}^n(\sup_{x\in[x_{k-1},x_k]}f(x)-\inf_{x\in[x_{k-1},x_k]}f(x))(x_{k}-x_{k-1}), $$ but this is just $$ (f(2)-f(0))\cdot \frac{2}{n}=\frac{8}{n}<\varepsilon. $$

Note that we were able to get all the way to the last line without knowing how fine the partition actually is.