Sigma-algebra requirement 3, closed under countable unions.

Question

The requirement for sigma-algebra is that.

1. It contains the empty set.

2. If A is in the sigma-algebra, then the complement of A is there.

3. It is closed under countable unions.

My question relates to 3. When we define the measure-function, it has to be disjoint countable unions, but not when we define a sigma-algebra.

I have tried to prove that if we require instead that the sigma algebra is closed under countable unions of disjoint sets, then it is closed under sets that are not disjoint. But this must be false, I think. Is there a way to prove this? That is I want to show that if I define a sigma algebra to be:

1. It contains the empty set.
2. If A is in the sigma-algebra, then the complement of A is there.
3. It is closed under countable unions of disjoint sets.

Then it is not nececarrily closed under countable unions if the sets are not disjoint. I guess this can be shown by creating a a set that satisfies the new definition, but is not closed under countable unions where what we take union over is not disjoint. But is it difficult to create a counterexample like this? If they don't exist then the definitions are equal, but I don't think they are?

Answer 1

Note that if $A_n$ is any family of sets, then $$ \bigcup_{n\in\mathbb N}A_n = \bigcup_{n\in\mathbb N}\bigl(A_n\setminus\bigcup_{k=0}^{n-1} A_k\bigr)$$ where the summands on the right-hand side are disjoint, and each of them is constructed from finitely many of the $A_i$s by a sequence of complements and finite unions.

So if you choose you can restrict yourself to requiring countable unions of disjoint set, if you also require that the algebra is closed under finite unions of arbitrary sets.

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A Dynkin system that is not a $\sigma$-algebra:

If you don't require arbitrary finite unions, what you get is not necessarily a $\sigma$-algebra. Consider for example the system of subsets of $\mathbb R$ consisting of $\varnothing$, $\{0,x\}$ for every $x\ne 0$, and the complements of these sets. It is closed under your proposed axioms, because the only nontrivial disjoint union there is to take is $\{0,x\} \cup \{0,x\}^\complement = \mathbb R$.

A small finite example is $$ \bigl\{\varnothing, \{0,2\}, \{0,3\}, \{1,2\}, \{1, 3\}, \{0,1,2,3\} \bigr\} $$

Answer 2

Assume $A_1,A_2,\ldots \in \Sigma$. Now define $$ B_i = A_i \setminus \bigcup_{i=1}^{n-1} A_i \text{.} $$ If $\Sigma$ is closed under finite unions (or intersections) and complements, it follows that $B_i \in \Sigma$. The $B_i$ are also obviously disjoint. And, you have for every $n$ $$ \bigcup_{i=1}^n A_i = \bigcup_{i=1}^n B_i \text{,} $$ and therefore also that $$ \bigcup_{i=1}^\infty A_i = \bigcup_{i=1}^\infty B_i \text{.} $$ Thus, if $\bigcup B_i \in \Sigma$ then also $\bigcup A_i \in \Sigma$.

It therefore doesn't matter whether you require that $\sigma$-Algebras are closed under disjoint countable unions or all countable unions, because if they're closed under disjoint countable unions, they are automatically closed under all countable unions. That equivalence requires that they are closed under finite unions though, so if you weaken the requirement to countable disjoint unions, you have to add either

• for $A,B \in \Sigma$, $A \cup B \in \Sigma$, or
• for $A,B \in \Sigma$, $A \cap B \in \Sigma$.

Otherwise, you end up with the definition of a Dynkin system, not a $\sigma$-Algebra.