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A set $E\subseteq \mathbb{R}$ is called 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?

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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 follwing 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

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  • $\begingroup$ A sigma algebra is closed to any countable unions, not necessary disjoint. $\endgroup$ – Mark Feb 18 at 10:42
  • $\begingroup$ 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 $\endgroup$ – gbox Feb 18 at 10:47

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