TITLE: Pull the teeth out of Lebesgue integration
QUESTION [11 upvotes]: In Lebesgue integration we usually approximate the function we want to integrate with step-functions on measurable sets. How much "power" do we take away if we require that the step functions are on intervals instead? What functions are left that are integrable?
I'm asking this because I want some integral to converge but I only know the values on $1_{(x_1,x_2)}$. Maybe we can get to Lipschitz functions that way?
Edit to be more specific.
Usually we define an integral for a positive function in this sense. First if we have
$$f = \sum a_i 1_{A_i} \text{ then } \int f \, d\mu = \sum a_i \mu(A_i)$$
Now if $f$ is a positive measurable function we then define
$$\int f \, d\mu := \sup \left \{\int g : g \leq f \text{ and $g$ is a simple function} \right \}$$
My question now is: what is left of the theory if we require the $A_i$ to be intervals instead of elements of the whole $\sigma$-algebra?
My apologies for the unclear question. I shouldn't ask questions in the middle of the night.

REPLY [9 votes]: There is a way to use step functions, and only step functions, to set up the Lebesgue integral due to Mikusinski. It requires something more subtle than a supremum. We say that a function $f$ is integrable if there exists a sequence $f_i$ of step functions such that $\sum \int |f_i|$ converges and such that $f(x) = \sum f_i(x)$ pointwise whenever $\sum |f_i|$ converges, and we define its integral to be $\int f = \sum \int f_i$.