Show that $f$ is Riemann integrable on $[a,b]$ and that $\int_{a}^b f=t-s$.

Question

Despite having looked at the many examples on here of proving (or disproving) that a function is Riemann integrable, I still can't quite wrap my head around how one would prove the following:

Suppose $a\leq s<t\leq b$. Define $f:[a,b]\rightarrow\{0,1\}$ by $$f(x)=\begin{cases} 1 & \text{if } s<x<t,\\ 0 & \text{otherwise.} \end{cases}$$ Prove that $f$ is Riemann integrable on $[a,b]$ and that $\int_a^b f=t-s$.

First I simply tried "unwinding" the definition that I am working with of what it means for a function to be Riemann integrable:

If $f$ is a bounded function on a closed, bounded interval, then $f$ is Riemann integrable if its lower Riemann integral equals its upper Riemann integral.

Ok, so the lower Riemann integral is defined as $$L(f,[a,b])=\sup_PL(f,P,[a,b])=\sup_P\sum_{j=1}^n(x_j-x_{j-1})\inf_{[x_{j-1},x_j]}f$$ and the upper Riemann integral is defined as $$U(f,[a,b])=\inf_PU(f,P,[a,b])=\inf_P\sum_{j=1}^n(x_j-x_{j-1})\sup_{[x_{j-1},x_j]}f$$ where both are taken over all partitions $P$ of $[a,b]$. So I want to show that $$L(f,[a,b])=t-s=U(f,[a,b]).$$ But I don't know how!

Is there a standard approach to solving these types of problems? I am certain I need to utilize some sort of $\epsilon,\delta$ argument here, but I am not very good with those types of proofs to begin with. I'm not really looking for an answer as much as I am looking for why the answer would be written the way it is i.e. what is your thought process when you see one of these and how does that process help you to a solution?

Answer 1

Hint: Suppose $a<s<t<b.$ Consider $U(f,P_n)- L(f,P_n)$ for the partition $P_n=\{a,s-1/n,s+1/n,t-1/n,t+1/n,b\}$ for large $n.$

Answer 2

Given a partition $P$ of $[a,b]$, then this partition can be of two types ($\underbrace{s\mbox{ and }t \in P}_{Type 1})$ or ($\underbrace{s \mbox{ or } t \not \in P}_{Type 2}$)

Then if $P$ is of Type1, $L(f,P,[a,b])=t-s$ because out of the interval $[s,t]$ we have $f(x)=0$ and hence $\inf f(x)=0$. In a same way you can show that $U(f,P,[a,b])=t-s$ too.

If $P$ is of Type2, you can see (and prove) that $L(f,P,[a,b])\leq L(f,P',[a,b])$ for any $P'$ of Type1. So, you can affirme (why?) that $\sup \limits_P L(f,P,[a,b])=\sup \limits _{P'}L(f,P',[a,b])$, where the first sup is taken over all $P$ and the second is taken over all $P'$ that are of type1.

Finaly, as we saw that the lower and upper sum using type1-partitions will always be the same constant, we conclude what we want!