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I like to know if there is a relationship between sets of Measure zero and $\sigma$-algebra. For instance, If a set $A$ or $A^{c}$ $\subset \mathbb{R}$ is of Lebesgue measure zero, what conclusion can we draw about the collection, say $\mathcal{C}$, of all the sets $A$.

Edit:

In other words if $\mathcal{C}$ is collection of all sets $A$, such that either $A$ or $A^c$ is of Lebesgue measure zero, can we say that $\mathcal{C}$ is a $\sigma$-algebra?

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Are you asking about the $\sigma$-ideal of Lebesgue null sets? –  t.b. Sep 13 '11 at 15:59
    
I have edited the question. hopefully, it is clear enough. Thanks. –  Kuku Sep 13 '11 at 16:08
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Yes, it's perfectly clear now. But I'm sure you can check this yourself! The empty set belongs to $\mathcal{C}$, the class $\mathcal{C}$ is closed under complementation, and for closure under countable unions you can distinguish two cases: either all sets are null (in which case the union is null) so or there is at least one co-null set (in which case the union is co-null). As I tried to point out, the more interesting concept is what you get if you consider the class $\mathcal{N}$ of all Lebesgue null sets. This is an example of a $\sigma$-ideal. –  t.b. Sep 13 '11 at 16:13
    
@Theo: Thanks for your response. Would you please post your recent comment as an answer, so I can accept it. –  Kuku Sep 13 '11 at 18:17
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1 Answer 1

up vote 6 down vote accepted

Let $(\Omega, \Sigma, \mu)$ be any measure space and let $\mathcal{C} \subset \Sigma$ be the class of sets $A \in \Sigma$ such that either $\mu(A) = 0$ or $\mu(A^c) = 0$. Then $\mathcal{C}$ is a $\sigma$-algebra:

  • Since $\mu$ is a measure we have $\mu(\emptyset) = 0$, so $\emptyset \in \mathcal{C}$.
  • By definition $A \in \mathcal{C}$ if and only if $A^c \in \mathcal{C}$.
  • If $A_1,A_2, \ldots \in \mathcal{C}$ then either $\mu(A_n) = 0$ for all $n$, so $\mu(\bigcup A_n) \leq \sum \mu(A_n) = 0$ and thus $\bigcup A_n \in \mathcal{C}$ or there is some $k$ such that $\mu(A_{k}^c) = 0$. But then $\Omega \smallsetminus \bigcup A_n = \bigcap A_{n}^c \subset A_{k}^c$, so $\bigcup A_n$ is conull and hence in $\mathcal{C}$. In both cases the union is in $\mathcal{C}$ and we are done.

Now let $\mathcal{N} = \{N \subset \Omega\,:\,\text{there is }E \in \Sigma \text{ such that }N \subset E\text{ and }\mu(E)=0\}$ be the class of negligible sets. Then $\mathcal{N}$ is a $\sigma$-ideal, that is to say

  1. $\emptyset \in \mathcal{N}$,
  2. If $N \in \mathcal{N}$ and $B \subset N$ then $B \in \mathcal{N}$,
  3. If $N_1,N_2,\ldots \in \mathcal{N}$ then $\bigcup N_n \in \mathcal{N}$.

Example: If $\Omega = \mathbb{R}$, $\Sigma = $ Borel sets and $\mu =$ Lebesgue measure then $\mathcal{N}$ is the class of Lebesgue null sets.

The structure of a $\sigma$-ideal plays a rôle in the theory of measure algebras and on a basic level in the context of completion of a measure space.

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thanks @Theo... –  Kuku Sep 13 '11 at 19:57
    
You're welcome. –  t.b. Sep 13 '11 at 20:04
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