4

A set $E\subseteq \mathbb{R}$ is called $m^*$-measurable if for all $A\subseteq \mathbb{R}$ $$m^*(A)=m^*(A\cap E)+m^*(A\cap E^c)$$

The set of all measurable sets is called Lebesgue sigma algebra

Does not sigma algebra has properties? this is a way to "build" Lebesgue sigma algebra from "another way" by measurable sets?

ZYX
  • 1
gbox
  • 13,645

1 Answers1

4

Yes, a $\sigma$-algebra has properties ! Let

$ \mathcal{L}= \{E \subseteq \mathbb R: m^*(A)=m^*(A\cap E)+m^*(A\cap E^c) \quad \forall A \subseteq \mathbb R\}.$

$ \mathcal{L}$ has the following properties (try a proof):

  1. $ \mathbb R \in \mathcal{L}$;

  2. $E\in \mathcal{L}$ implies $\mathbb R \setminus E \in \mathcal{L}$;

  3. if $(E_j)$ is a sequence in $ \mathcal{L}$, then $\bigcup E_j \in \mathcal{L}.$

This shows that $ \mathcal{L}$ is a $ \sigma$-algebra.

James
  • 1,079
  • 7
  • 26
Fred
  • 78,422
  • A sigma algebra is closed to any countable unions, not necessary disjoint. – Mark Feb 18 '19 at 10:42
  • So why did my lecture said that if Carathéodory criterion is met, it is a Lebesgue sigma algebra, is should be just a sigma algebra – gbox Feb 18 '19 at 10:47
  • Probably because they were just going to use the Lebesgue measure since this is the most common one in introductory integration and measure theory classes. – NDewolf Mar 25 '20 at 19:42
  • I agree with @NDewolf ... The "Lebesgue sigma-algebra" is the result of the OP definition in the case where $m^$ is the Lebesgue outer measure. For other outer measures, I may call it "$m^$-measurable sets". – GEdgar Mar 01 '24 at 13:58