In Real Analysis by Royden (4th) edition, Royden has the following proof that Riemann integrability implies Lebesgue integrability for a bounded function $f$ defined on a closed, bounded interval $[a,b]$.
I am not understanding the logic here. He claims that because each step function is a simple function the first equality implies the second. However, even though every step function is a simple function, it is not true the other way around. That is, there are simple functions that are not step functions, and the second inequality is therefore taking the $\sup$ over this larger class of functions. So, theoretically, couldn't the second equality yield a different result than the first?
We have $\sup_1 \le \sup_2 \le \inf_2 \le \inf_1.$ Since $\sup_1 =\inf_1,$ these numbers are all the same. (Here the subscript $1$ refers to your first line, and the subscript $2$ refers to you second line.)