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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.

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    The Borel sigma-algebra is defined to be the sigma-algebra generated by open sets, so intervals are already enough to specify Lebesgue measure. You get the same theory.2011-02-20
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    @Qiaochu: Really? Maybe I don't understand you answer, but I have a step function of the form $\sum a_i 1_{(x_i, x_{i + 1})}$ instead of on a general measurable set, can all positive integrable functions be given by limits of such functions? That is how we usually define this for positive functions.2011-02-20
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    I'm puzzled: Approximate a function by characteristic functions of intervals in what sense? Uniformly? Then you get regulated functions and the regulated integral. Could you be more specific, please?2011-02-20
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    Step functions on intervals are simply piecewise constant functions, right? So doesn't that reduce your integral to the Riemann integral, since by definition the Riemann integral of $f(x)$ is just the common value of the sup over all integrals of piecewise constant functions $g(x) \leq f(x)$, and the inf of all integrals of piecewise constant functions $g(x) \geq f(x)$.2011-02-20
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    @Jonas: yes, that is precisely what I'm claiming. It suffices to show that bounded measurable sets can be approximated by finite unions of intervals (in the sense that their symmetric difference can be made arbitrarily small). This is a good exercise and is a precise form of Littlewood's first principle. (When you say "limit" I assume you mean in the L^1-norm.) @Greg: I don't see how this follows.2011-02-20
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    @Theo: I have editted the question a bit to explain what I mean. @Qiaochu: Thank you, I will think about it. I was not aware of Littlewood's principles.2011-02-20
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    @Qiaochu: For instance the function f(x) = 1 if x irrational, 0 if x rational defined on [0,1] is going to have integral 0 by Jonas' scheme (or won't be integrable, depending how Jonas wants to define things), whereas the Lebesgue integral ought to be 1. Right?2011-02-20
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    Thanks for the clarification. The problem is that this won't give you a linear map without some hypotheses about the nature of convergence (and as I said, if you require uniform convergence you get the regulated integral - a slight extension of the Riemann integral). Moreover, it's easy to see that your process won't give you the usual integral. Take for example the characteristic function of a modified Cantor set. The only step function lying below it is the zero function.2011-02-21
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    @Theo: Okay, I see. So basically this way we can still integrate all continuous functions (a.e.) on bounded intervals (if assuming uniform convergence). Anyone, please post an answer.2011-02-21
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    During the time between opening the question and posting my answer, there were comments I didn't notice. In particular, I inadvertently duplicated examples of both Greg O. and Theo Buehler.2011-02-21
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    @Greg: I see. I misinterpreted Jonas' implicit definitions.2011-02-21

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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.

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    @Jonas T: I was thinking more in terms of the ability to approximate in $L^1$ (with Lebesgue measure), after the work has already been done to define the integral, rather than actually defining the Lebesgue integral of a measurable $f$ in terms of a limit of integrals of step functions. I'm not sure what the use of the latter approach would be, and off hand I don't know how it would be done rigorously without effectively building up all of the theory with simple functions as a side effect, thus leading to a lot of redundancy. But these are naïve thoughts.2011-02-21
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    @Jonas: well, no. I was under this impression in the comments, but the problem is that to define what it means for a function to be the L^1-limit of step functions you already need a definition of the Lebesgue integral.2011-02-21
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    @Jonas: Sorry, I edited out my question. For the rest, I was asking if we could define the integral as an $L^1$ limit. This question came from how we define stochastic integrals.2011-02-21
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    @Jonas T: You're welcome. (My last comment was an answer to a question that was edited out of a comment. Perhaps I should delete it?)2011-02-21
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    @Jonas: No, I would not delete it, the continuity of the comments would become even worse then...2011-02-21
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    Update to the naïve thoughts in my first comment: Qiaochu's answer indicates how it can be done.2011-02-21
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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 $\int f = \sum \int f_i$.

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    Interesting. Do you happen to have a reference?2011-02-21
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    @Jonas: http://books.google.com/books?id=tYrG-JlqoQQC&printsec=frontcover&dq=hilbert+spaces+with+applications+mikusinski2011-02-21