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I just wanted to clarify the difference between the Algebra and $\sigma$-algebra:

Algebra: If $A_1, A_2 \ldots $ are in $\mathscr A$, then $\bigcup_{i = 1}^{n} A_i \in \mathscr A$
$\sigma$-Algebra: $A_1, A_2 \ldots $ are in $\mathscr A$, then $\bigcup_{i = 1}^{\infty} A_i \in \mathscr A$

Notice the difference is just between $n$ and $\infty$. Let's use this following as illustration:

(1) Let $X$ be a set; let $A, B, C, D \subset X$

(2) Let $\mathscr A_1 = \{\emptyset, A, A^c, X\}$ and $\mathscr A_2 = \{\emptyset, B, B^c, X\}$ be $\sigma$-algebras of $X$

(3) Then $\mathscr A_1 \cup \mathscr A_2 = \{\emptyset, A, A^c, B, B^c, X\}$

(4) Here, $\mathscr A_1 \cup \mathscr A_2$ is algebra because $\emptyset \cup A \cup A^c \cup B \cup B^c \cup X = X$

(5) But $\mathscr A_1 \cup \mathscr A_2$ is not a $\sigma$-algebra because in order to qualify, any pick (or any length) of union of elements of $\mathscr A_1 \cup \mathscr A_2$ has to be in $\mathscr A_1 \cup \mathscr A_2$ also.

(6) Here, since $A \cup B$ is not in $\mathscr A_1 \cup \mathscr A_2$, therefore $\mathscr A_1 \cup \mathscr A_2$ is not $\sigma$-algebra

(7) However, if $A, B \subset X$ only, then $\mathscr A_1 \cup \mathscr A_2$ is $\sigma$-algebra since $A \cup B = X$ and $ X \in \mathscr A_1 \cup \mathscr A_2$.

Please let me know if I am wrong, especially (5) to (7). Thank you very much for your time and effort.

A.Magnus
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  • Note that any finite algebra is a $\sigma$-algebra - you'll need to look at infinite algebras to see the difference. – Mario Carneiro Jan 28 '15 at 14:11
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    Your example is not an algebra because $A\cup B\notin{\scr A}_1\cup{\scr A}_2$. – Mario Carneiro Jan 28 '15 at 14:13
  • That is actually what I don't understand the difference between algebra and sigma-algebra. What do you mean by infinite algebra? – A.Magnus Jan 28 '15 at 14:18
  • An algebra is closed under finite unions (it is sufficient to consider only binary unions, i.e. $A\cup B\in{\scr A}$ for any $A, B\in{\scr A}$), a $\sigma$-algebra is closed under countable unions. I believe you are missing closure under complementation as well, though. – Mario Carneiro Jan 28 '15 at 14:23
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    The set of all finite or co-finite subsets of $\Bbb N$ is an algebra that is not a $\sigma$-algebra. – Mario Carneiro Jan 28 '15 at 14:25
  • Can you explain the difference (using example perhaps) between FINITE versus COUNTABLE union? – A.Magnus Jan 28 '15 at 14:30
  • A countable union is a union of a countably infinite collection of sets. For example, in my example the set of even numbers is ${2}\cup{4}\cup{6}\cup\dots=\bigcup_{n=1}^\infty{2n}$ where each singleton set is finite, so if it were a $\sigma$-algebra then the set of all even numbers would be in my algebra. But it's not finite, and its complement is the set of all odd numbers which is not finite either, so it's not in my algebra. Thus it's not a $\sigma$-algebra. – Mario Carneiro Jan 28 '15 at 14:37

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(4) is incorrect: An algebra is closed under finite unions, but $A \cup B \notin \mathscr A_1 \cup \mathscr A_2$, necessarily. This means that $\mathscr A_1 \cup \mathscr A_2$ is not an algebra and hence can't be a $\sigma$-algebra.

If you have two algebras, their union isn't necessarily an algebra as you can deduce from the answers in your other question here.

In (7) you say that $A \cup B = X$, but you haven't stated that anywhere above, all you've said is that $A, B \subseteq X$, which doesn't guarantee their union to be $X$.

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Robert Cardona
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