My textbook has a lot of definitions that look more or less the same thing to me so excuse my ignorance.
It first defines a simple function as a function that can be written as
$$f=\sum_{k=1}^na_k\chi_{A_k}$$
where $A_1,..,A_n$ are disjunct pairwise. Then it defines the integral of $f$ with respect to the measure $\mu$ in the usual way.
Now, if the function $g$ isn't simple, then it defines it's integral (wrt. $\mu$) as the supremum of all integrals of simple functions for which $f(x) \leq g(x)$ for all $x$.
Now my question is, if $\mu$ is the Lebesgue measure, is this the same as the supremum of lower Darboux sums? On one hand it would make sense because if the Riemann integral requires that the lower and upper Darboux sums converge to the same number, if the Lebesgue integral only required that the lower sums had a supremum, it would explain why more functions are Lebesgue integrable than Riemann integrable. On the other hand, I feel like this fact would be mentioned in the text.
What exactly am I missing here? Again, sorry if the question doesn't make sense. I realize that this area of mathematics is full of subtleties that I don't yet understand.
I believe there is one key difference. Namely, the sets $A_k$ in the Lebesgue integral do not have to be intervals, they just have to be Lebesgue measurable. However, in a lower Darboux sum, your sets are restricted to intervals. To see why this matters, consider the function $$ f(x)=\begin{cases}0\text{ if }x\in\mathbb{Q}\\1\text{ if }x\not\in\mathbb{Q}\end{cases} $$ on the interval $[0,1]$. You can show that any lower Darboux sum will evaluate to $0$, as there is a rational number in any nontrivial interval of $[0,1]$. However, the Lebesgue integral evaluates to $1$, as we can take $A_1=\mathbb{Q}$ and $A_2=[0,1]\setminus \mathbb{Q}$.