A new notion of measurable set

Question

The idea behind the Riemann integral is to approximate functions by step functions, for which there is a natural notion of integral. More specifically, we consider all step functions that lie either entirely above or entirely below our function, assign integrals to each of the step functions in these families, and then consider respectively the infimum or supremum of these two families.

Critically, if the two numbers match, we say that the function is Riemann integrable. My question is about the definition of Lebesgue measurable, and why it doesn't seem to follow the same idea.

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We can define the outer measure of a subset $S$ of $\mathbb{R}$ as the infimum of the 'sizes' of countable unions of disjoint intervals that contain $S$. My instinct, now, might be to define some sort of 'inner measure' as the supremum of the 'sizes' of countable unions of disjoint intervals that are contained within $S$. I would like to say that a set is measurable if these two numbers agree.

A little thought reveals that this is a very restrictive definition. The set of irrational numbers within $[0,1]$, for instance, is not measurable in this sense. Indeed, I think that any such 'measurable' set must be a countable union of disjoint intervals and discrete points.

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I now have two questions:

Question 1: is the characterisation above correct? Is a set which is 'measurable' in the sense I've defined it simply a countable union of disjoint intervals and discrete points? Edit: The answer is "no" (see Michael's answer below); the Cantor set is a counterexample. I ask instead: what is the correct characterisation of such 'measurable' sets?

Question 2: could we have known, ahead of time, that this idea would not work? It seemed sensible to me at first, by analogy with the Riemann integral, to define measurable in the above way, and it was only after exploring the consequences that I saw it didn't work. Is there some intuition or understanding that would have led me to discard the idea immediately?

Answer 1

Regarding Question 1: For measurable sets, your given outer approximation works. So take the Cantor set, a measure zero set that does not consist of discrete points. You cannot fit in any nontrivial interval, so its inner approximation assigns measure zero to it, as does the outer approximation. So you have a counterexample.

Regarding Question 2: First, it is not really clear what "ahead of time" means here. Certainly, it doesn't take too long to figure out the problems. But if your knowledge comes from the side of Riemann integration, you could have seen that the indicator function of many sets is not Riemann integrable either. Or you might try to prove a property such as additivity and quickly come op with a counterexample such as the one you gave.

Answer 2

To answer question $1$, note that any set $X\subseteq\mathbb{R}$ has a maximal countable disjoint set of intervals contained in it - namely, the set of connected components of $int(X)$ (the interior of $X$). The sum of the sizes of these is just the measure of $int(X)$ in the usual sense, and this is clearly the "inner" measure of $X$ in your sense. This gives us:

A set is "measurable" in the sense of the OP if $X\setminus int(X)$ is null.

That is, Michael's guess in the comment below his answer is correct.

Answer 3

Let $m^o$ denote Lebesgue outer measure. For $A\subset \Bbb R$ we can define the Lebesgue inner measure $m^i(A)=\sup \{m^o(C): C=\overline C\subset A\}.$ And define $A$ to be Lebesgue measurable iff $m^i(A)=m^o(A).$

The Cantor set has measure $0.$ But there are sets often called "fat Cantor sets" that are closed, have empty interior and no isolated points, and have positive Lebesgue measure.

The idea of Lebesgue integration is: Rather than partitioning the domain of the function $f$ into intervals, partition the range into intervals $\{I_j\}_{j\in J}$ (where $J$ is finite ) and approximate $\int f(x)dx$ by $\sum_{j\in J}\;y_j\cdot m(f^{-1}I_j)$ where each $y_j\in I_j.$