4

I am trying to work out what I think (if successful) is the "most natural" construction of Lebesgue measure, using transfinite recursion. There are already multiple papers on this, including two papers by Deiser mentioned in this answer, but it seems they still use various inner/outer approximations and estimations. I would like a construction that is "$\varepsilon$-free".

For simplicity let's work with $[0,1)$ instead of $\mathbb{R}$. We start with the finite premeasure on the collection of finite unions of left-closed-right-open subintervals of $[0,1)$; that this is indeed a premeasure can also be shown using transfinite induction. The basic idea of extending this premeasure is very simple, and comes from the $\pi$-$\lambda$ theorem: if the measure $\mu$ has already been defined on $A,B$ where $A\subseteq B$, then we are forced to define $\mu(B\setminus A)=\mu(B)-\mu(A)$. Similarly, if $\mu$ has been defined on $\{A_n:n<\omega\}$ and they are disjoint, we must have $\mu(\bigcup_n A_n)=\sum_n \mu(A_n)$.

So we may want to stratify the Borel sets, equivalently the $\lambda$-system generated by the $\pi$-system of left-closed-right-open intervals, into a hierarchy of length $\omega_1$, and define the Lebesgue measure recursively. But it seems cumbersome to check that this is well-defined and that it is a measure, because we are adding too many sets at each step. An alternative strategy is to add new sets one by one. But then we won't have a hierarchy, but a premeasure defined on each countable fragment of the Borel $\sigma$-algebra. To summarize, below is my proof plan:

  1. Start with the premeasure defined on the algebra of set on $[0,1)$ consisting of finite unions of left-closed-right-open intervals.
  2. Whenever we have a finite premeasure $\mu$ defined on an algebra of set $\mathcal{E}$, and $\{C_n:n<\omega\}$ is a sequence of disjoint sets from $\mathcal{E}$, we can let $C=\bigcup_n C_n$ and consider the algebra of sets generated by $C$ over $\mathcal{E}$; a set in this new algebra is of form $A\cap C\cup B\cap C^c$ where $A,B\in\mathcal{E}$. Extend $\mu$ to this new algebra by defining $$\mu(A\cap C\cup B\cap C^c)=\sum_n\mu(A\cap C_n)+\mu(B)-\sum_n \mu(B\cap C_n)$$ We need to check that this is well-defined, and that $\mu$ remains a premeasure. I think I figured out well-definedness, but haven't checked the latter (it must be true, of course, but I'm not sure if it follows from basic set manipulations).
  3. Call an algebra of set on $[0,1)$ good if it is the last one of a sequence $(\mathcal{E}_i)_{i\leq\alpha}$, where $\alpha$ is some countable ordinal, $\mathcal{E}_0$ is the algebra in 1, $\mathcal{E}_{i+1}$ is obtained from $\mathcal{E}_i$ by the procedure in 2, and union is taken at limit stage. Use 2 to show that for any countably many Borel sets, there is a good algebra of sets containing them, together with a premeasure defined on the algebra. (There seems to be a major issue here. There is no reason why the property of being a premeasure is preserved at limit stage)
  4. Argue that the premeasure in 3 is unique.
  5. For an arbitrary Borel set $A$, define its measure by choosing any algebra of set containing $A$ and using the premeasure on it.

There are many details to check, but does the overall strategy look right? Is this approach already written somewhere?

Kushi
  • 425
  • I have just found this paper in this question. I think it might be of interest. They use lattices and basically follow the Borel hierarchy, and obtain regularity results. I have just skimmed it, though. You might be interested in this related question of mine (although it seems that it is not necessary for uniqueness). – Keplerto Aug 14 '24 at 15:18
  • Sorry for commenting before reading the whole post. The issue that you have with a limit stage $\lambda$ seems to be that, for a set $C$ born at stage $\alpha < \lambda$, there may be a sequence $A_n$ of sets born at stages $\beta_n$ increasing to $\lambda$ such that $C$ is the disjoint union of the $A_n.$ The fact that $\mu$ is a premeasure on each stage before $\lambda$ does not guarantee that $\mu(C) = \sum_n \mu(A_n).$ A natural way to solve this problem is to strengthen the induction hypothesis. For example, we may prove that the measure is regular, and can be approximated by, for example – Keplerto Sep 05 '24 at 22:24
  • by open sets from the outside. This seems to be the type of argument that you do not want to use. But if you don't prove directly that the measure is regular, you will have to do it later on when you need that property. I think that adding elements one by one, as you are doing, is not really that different from the Borel hierarchy construction – Keplerto Sep 05 '24 at 22:32

0 Answers0