Why is the naive recursive approach to defining Lebesgue measure not satisfactory?

Question

As I understand it, the spirit of the Lebesgue measure is:

An interval is measurable and the measure of an interval is the absolute difference between its endpoints. The measure of a countable union of disjoint measurable sets is the sum of their measures. Only sets that this recursive definition "reaches" are measurable.

I realize that the problem is that in some sense this tells you "a set is measurable if" without providing a very precise "...and only if". But I can't seem to really convince myself that this is a problem. The basic idea is that this definition provides an algorithm for determining the measure of a set. The algorithm either does or doesn't apply to a given set. Are there sets such that their measurability is ambiguous under the above definition?

Answer 1

Your definition is essentially fine, but incomplete.

You need to indicate that the union you are taking is disjoint, and also need a clause stating that if $A$ is a measurable subset of an interval $I$, then the measure of the complement of $A$ in $I$ is the measure of $I$ minus that of $A$. Once you add this clause your "naive" definition indeed provides the right measure for all Borel sets. The recursion takes $\omega_1$ steps, meaning that if $\alpha$ is a (possibly infinite, but) countable ordinal, then running your definition for only $\alpha$ many stages will fail to assign measure to some Borel sets.

To be precise: We can call $A$ a level $0$ set iff it is an interval, for $\alpha$ countable, call $A$ a level $\alpha+1$ set iff it is not a level $\alpha$ set, but it is the countable union of sets $A_n$ of level $\alpha$ or the complement of a level $\alpha$ set, and call $A$ a level $\gamma$ set, for $\gamma$ a countable limit ordinal, iff it is not a level $\beta$ set for any $\beta<\gamma$, and there is a sequence $\gamma_n$ of ordinals smaller than $\gamma$, and for each $n$ there is a set $A_n$ of level $\gamma_n$, with $A=\bigcup_n A_n$. Our definition assigns measures to all sets of level $\alpha$, for any countable $\alpha$, and for each $\alpha$ there is at least one set of level $\alpha$, so we cannot stop before that stage. On the other hand, the collection of all sets of countable level is closed under the clauses of your definition.

(In descriptive set theory we actually use something very close to your approach to describe the Borel complexity of a set. This approach, of assigning a level of complexity to sets, allows us to prove statements about Borel sets by transfinite induction, and is indeed very useful. In practice, however, it is quite rare to happen upon a Borel set whose level is beyond, say, $5$.)

To see the need for the extra clause regarding complements: We want singletons to be measure zero. This is perhaps fine if we are liberal with our interpretation of the word "interval". But without the extra clause, not even the Cantor set gets assigned a measure.

Additional clauses will be needed if you want to reach all Lebesgue measurable sets. Perhaps the simplest is to say that if $A$ is measurable and has measure zero, then any subset of $A$ is also measurable with measure zero, and that if $A$ is measurable and $B$ has measure zero, then $A\cup B$ is measurable and has the same measure as $A$.

Of course, there are difficulties with this approach. For instance, one needs to verify that it is self-consistent, meaning that if a set $A$ is witnessed to be of level $\alpha$ in two different ways, we need to prove that the two "witnesses" assign to $A$ the same measure. Similarly for the additional clause I mentioned in the paragraph above. One also needs to verify that this approach captures all measurable sets. (I did not require that the unions verifying a set is level $\alpha$ be disjoint, although I required this when it came to assigning measures. Some additional care is needed to deal with this detail.)

Answer 2

There are two problems that I can see:

1. As goblin mentions in the comment, only sets that are countable union of disjoint intervals are "measurable" according to your definition. Unfortunately, this is not enough. For example, the measure of the Cantor set or the fat Cantor set remain undefined.

2. You would still need to show that your definition is consistent. That is, if you can write a set $A$ as a countable union of disjoint intervals in two different ways, both decompositions lead to the same value for the measure of $A$.

Edit:

If you change your definition so that the complement of measurable set $A$ has measure $1$ minus the measure of $A$ (assuming your space is the unit interval), then your family of a measurable sets will be extended and sets like the Cantor set will be covered. However, there will still be two difficulties:

3. You would still need to show that the intersection of two measurable sets is measurable. (I am not confident that this would be the case.) This property is crucial in applications, for instance in the definition of conditional probability, or in the proof of the continuity of the measure.

4. You would still have a highly non-trivial task of showing the consistency of your definition.

Answer 3

There are some technical nitpicks with your definition, such as you want a countable disjoint union before you can just sum the measures of the uniees. If you fix those, it seems likely that you end up with the standard Borel measure on $\mathbb R$.

The problem with using this as your default one-size-fits-all measure is that it doesn't satisfy "squeeze" theorems such as

Let $A$ be a set and $L$ some number. Assume that for every $\varepsilon>0$ there exist $B\subseteq A$ and $C\supseteq A$ such that $\mu(B)>L-\varepsilon$ and $\mu(C)<L+\varepsilon$. Then $A$ is measurable with measure $L$.

In fact, the Lebesgue measure can be characterized as the unique "least defined" (that is, with the smallest possible domain) measure that extends the standard Borel measure and satisfies this squeeze property.

Answer 4

This should work fine after, as some people suggested, you add complementation and, I am suggesting, also allow changes modulo null sets. More explicitly, you can first show that you get a well defined countably additive set function on all G-delta sets that coincides with length on intervals and then extend this function to all sets which equal a G-delta set modulo a null set where you can define null sets of reals in the usual way.