I have $RI[a,b]$ as the set of all Riemann Integrable functions and $d_1(h_\alpha, h_\beta)$ where $$h_\alpha(t)= \begin{cases} \alpha & t\in\{a,b\} \\ 0 & a<t<b \end{cases}$$
Through sketches I can see that calculated $d_1(h_\alpha, h_\beta) = 0$ but I ideally want to prove this with something more written down, I thought I could prove this using Riemann sums, so I have the formula below and would want it equal to zero as shown: $$\sum_{j=1}^n f(x_j)\Delta x = 0$$
I'm struggling with translating the information I have into this format that I could use for the sum, any help would be great, thanks!
Given that the function you are integrating is positive and you want to prove the integral is $0$ through Riemann sums. I'll prove it using upper Riemann sums tend to $0$ for sufficiently fine partitions (That would prove integrability and convergence). The key here being that they only differ on the ends of the interval.
So let $\varepsilon>0$ and $\alpha,\beta\in\mathbb{R}$ Given that we want to prove that $$\int_a^b |h_\alpha(x)-h_\beta(x)|dx=0$$ Take the following partition $x_0=a<x_1=\frac{\varepsilon}{2|\beta-\alpha|}<\ldots<x_{n}=b-\frac{\varepsilon}{2|\beta-\alpha|}<x_{n+1}=b$. Then our upper Riemann sums ($S_\pi$) are: $$|S_\pi| = \left|\sum_{k=0}^n \sup_{x\in[x_k,x_k+1]}|h_\alpha(x)-h_\beta(x)| \Delta x\right| = |\beta-\alpha|\cdot \frac{\varepsilon}{2|\beta-\alpha|} + |\beta-\alpha|\cdot\frac{\varepsilon}{2|\beta-\alpha|} = \varepsilon $$ Given that we can do this for any $\varepsilon >0$, then it follows that $S_\pi\to_{|\pi|\to0} 0 $