I am currently studying Lebesgue measure theory, and I had some questions about the definition of the Lebesgue measure.
So, to start off, we are looking for a well-defined, positive function $m$ acting on real sets, satisfying
- For every interval $I$, $m(I)=|I|$ ($I$ is the interval's length).
- $m(E+x_0)=m(E)$ (invariance to shifts of the set).
- Sigma additivity: $m(\bigcup_n E_n)=\sum_n E_n$, for disjoint $E_n$.
The outer measure is defined as
$$m^*(E)=\text{inf}\space \{\sum_n |I_n| : E\subset\bigcup I_n \}$$
And it satisfies all requirements, except sigma-additivity (it only satisfies sub-sigma-additivity). It can also be proven that there is no function satisfying all requirements. Therefore, the requirement of acting on all real sets is dropped, and we define the Lebesgue measure as the outer measure acting only on "Lebesgue measurable" sets. These are sets $E$, for which for all $A\subset \mathbb{R}$,
$$m^*(A)=m^*(A\cap E)+m^*(A\cap E^c)$$
I have two questions. Firstly, is my understanding of how the Lebesgue measure defined correct? Secondly, if it is, can it be shown that Lebesgue measurable sets are the only real sets for which the outer measure is sigma-additive?
I really don't know how to even think about this intuitively, and it might be beyond my competence (or very simple?), but thought I'd ask.
Best Answer
Your thinking seems correct.
Non-measurable sets can be sigma-additive. Let's consider a Vitali set $V$ (or any other non-measurable set on $[0,1]$) and now consider $\{V+2n: n \in \mathbb{N}\}$. It's easy to see that these disjoint sets are sigma-additive. in fact you could change it a little bit (appropriately contracting the sets) and get a finite-measure sum as well.
In fact more generally: if we have $dist(A_i, A_j) > M$ for every pair $i \neq j$ and some constant $M>0$ independent from $i, j$, then $A_i$ are sigma additive in Lebesgue measure.
Now an interesting thing is that if you drop the Axiom of Choice, then you can construct a measure stisfying your conditions which is defined on every subset of $\mathbb{R}$. So the existance of non-measurable sets requires the Axiom of Choice. See here: https://en.wikipedia.org/wiki/Solovay_model.