Pull the teeth out of Lebesgue integration

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.

Terminologically speaking, step function usually means linear combination of characteristic functions of intervals, whereas simple function is used for functions taking on finitely many values.

If you're talking about Lebesgue measure, you can approximate arbitrary $L^1$ functions with step functions in $L^1$ norm. It's just that in defining the Lebesgue integral, it's easiest to do so first for arbitrary simple functions, because then all measurable functions can be approximated in a particularly nice way. In particular, if $f$ is nonnegative, then $\int f$ can be defined to be the supremum of the integrals of nonnegative simple functions dominated by $f$. This would not work for step functions, as the characteristic function of the irrationals in $[0,1]$ shows; or, less trivially, the characteristic function of a fat Cantor set. Thus, your proposed change in definitions would break down for relatively nice Borel functions.

However, the problem of $L^1$ approximation by step functions could be settled if you could find, for each finite measure set $E$ and for each $\varepsilon\gt0$, a finite union of intervals $F$ such that $\int |\chi_E - \chi_F |\lt \varepsilon$. That is, you want the measure of the symmetric difference of $E$ and $F$ to be less than $\varepsilon$. That this is true for Lebesgue measure on the line is (a version of) one of Littlewood's 3 principles.

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 $\int f = \sum \int f_i$.