1
$\begingroup$

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.

$\endgroup$
7
  • $\begingroup$ Note that any finite algebra is a $\sigma$-algebra - you'll need to look at infinite algebras to see the difference. $\endgroup$ Jan 28, 2015 at 14:11
  • 1
    $\begingroup$ Your example is not an algebra because $A\cup B\notin{\scr A}_1\cup{\scr A}_2$. $\endgroup$ Jan 28, 2015 at 14:13
  • $\begingroup$ That is actually what I don't understand the difference between algebra and sigma-algebra. What do you mean by infinite algebra? $\endgroup$
    – A.Magnus
    Jan 28, 2015 at 14:18
  • $\begingroup$ 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. $\endgroup$ Jan 28, 2015 at 14:23
  • 1
    $\begingroup$ The set of all finite or co-finite subsets of $\Bbb N$ is an algebra that is not a $\sigma$-algebra. $\endgroup$ Jan 28, 2015 at 14:25

1 Answer 1

1
$\begingroup$

(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 $X$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .