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I am looking for (the name of) a class of functions from $\mathbf{R}^2$ $(\mathbf{R}^n)$ to {0, 1} that are integrable.

Let $f$ be in this class and $E$ be the set of all points where $f$ is equal to $1$. Then for all $x\in E$, the following holds for all $r>0$: $$ \operatorname{area}( B(x, r) \cap E ) > 0. $$ The same holds for the points where $f$ is $0$.

That is, $f$ can not arbitrarily be changed on a null set and still remain in this class. The goal is to be able to talk about point-wise values of solutions to optimization problems involving integrals.

Does this class (or a related one) have a name?

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On the Wikipedia page for the Lebesgue density theorem (which is very closely related to what you ask about) it is noted that if $E$ and its complement both have positive measure then there are always points where the "approximate density" is strictly between $0$ and $1$. It is then always possible change $E$ by either including or removing such points. –  Jack Huizenga Jan 30 '13 at 13:28
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You might want to reconsider whether you're sure that ${\rm area}(B(x,r) \cap E) > 0$ for all $x \in E$ and $r > 0$. That's not even true for Riemann integrable functions into $\{0,1\}$. –  Nik Weaver Jan 30 '13 at 13:30
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I don't know a name for this property, but note that any Lebesgue measurable set $F\subset\mathbb{R}^n$ has a subset $E$ (the set of its points of density $1$) with the even stronger property that |B(x,r)∩E|=|B(x,r)|(1+o(1)), and |F\E|=0. (see en.wikipedia.org/wiki/Lebesgue_density_theorem ). –  Pietro Majer Jan 30 '13 at 13:32
    
Nik: Right, the set of all Riemann integrable functions is too large for my purposes. –  Ben Jan 31 '13 at 0:04
    
I do not think you can go beyond step functions –  Guido Kanschat Sep 9 '13 at 7:01

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