18
$\begingroup$

Here is another very basic analysis problem but that puzzles me:

Find an example of set $X$ and its two $\sigma$-algebras $\mathscr A_1$ and $\mathscr A_2$, such that $\mathscr A_1 \cup \mathscr A_2$ is not $\sigma$-algebra.

To me at least, this question looks counter-intuitive since the union of two sets gives the resulting set larger number of elements, thus won't affect its $\sigma$-algebra status.

Please help and thank you for your time and effort.

$\endgroup$
1
  • 1
    $\begingroup$ The problem becomes much harder if we change it to be: "Find an example that the union of two sigma algebras is an algebra but not a sigma algebra." $\endgroup$
    – High GPA
    Aug 17, 2023 at 7:49

3 Answers 3

32
$\begingroup$

take $X := \{a,b,c\}$ and $A_1 := \{ \{a\}, \{b,c\}, \emptyset, X\}$, $A_2 := \{ \{b\}, \{a,c\}, \emptyset, X\}$ and show that $A_1 \cup A_2$ is not a $\sigma$-algebra

$\endgroup$
7
  • $\begingroup$ Thanks & I like your solution. But forgive my slowpoke mind, I didn't see it fails the conditions. $X$ and $\emptyset$ are there, all complements are there, the union of all is the $X$, and the intersection of all is there, $\{a\} \cap \{b, c\} \cap \{b\} \cap\ \{a, c\} = \emptyset$. I am sure I missed something. What is that? $\endgroup$
    – A.Magnus
    Jan 27, 2015 at 18:15
  • 3
    $\begingroup$ @A.Magnus $\{a\}\cup\{b\}=\{a,b\}$ is not there. $\endgroup$
    – drhab
    Jan 27, 2015 at 18:21
  • $\begingroup$ @drhab : When we talk about the union (or intersection) of elements in the conditions, is that union (or intersection) of two elements at a time? Or all elements together? Because if it's about all elements together at the same time, then $\{a\}\cup \{b,c\}\cup\{b\}\cup \{a,c\}=\{a, b, c\}$, which is $X$. Did I miss again? $\endgroup$
    – A.Magnus
    Jan 27, 2015 at 18:30
  • 3
    $\begingroup$ Let $\mathcal A:=\{\emptyset,\{a\},\{b\},\{b,c\},\{a,c\},X\}$ i.e. the union of the two sigma-algebras mentioned in this answer. If $\mathcal A$ would be a sigma-algebra (wich is not the case) then it should satisfy: if $A_n\in\mathcal A$ for $n=1,2,\dots$ then $\bigcup_{n=1}^{\infty}A_n\in\mathcal A$. However, if we take $A_n=\{a\}$ for $n$ even and $A_n=\{b\}$ for $n$ odd then we get $\bigcup_{n=1}^{\infty}A_n=\{a,b\}\notin\mathcal A$. $\endgroup$
    – drhab
    Jan 27, 2015 at 18:49
  • $\begingroup$ @drhab : So am I wrong if I write like these: $\bigcup_{n=1}^{6} := A_1 \cup A_2 \cup A_3 \cup\ A_4 \cup A_5 \cup A_6 = \emptyset \cup \{a\} \cup \{b\} \cup \{b,c\} \cup \{a,c\} \cup X = X$? Do let me know & thanks. $\endgroup$
    – A.Magnus
    Jan 27, 2015 at 19:31
3
$\begingroup$

$\{\emptyset,[0,3/4),[3/4,1),[0,1)\}$ and $\{\emptyset,[0,1/2),[1/2,1),[0,1)\}$ are $\sigma-$algebras but their union $\{\emptyset,[0,3/4),[3/4,1),[0,1/2),[1/2,1),[0,1)\}$ is not. $[0,3/4)\ minus \ [1/2,1)=[1/2,3/4)$ is not in the union.

$\endgroup$
2
$\begingroup$

Let it be that $X=A\cup B=U\cup V$ with $A\cap B=\emptyset=U\cap V$. Then $\mathcal{A}=\left\{ \emptyset,A,B,X\right\} $ and $\mathcal{V}=\left\{ \emptyset,U,V,X\right\} $ are both $\sigma$-algebras. Is $\mathcal{A}\cup\mathcal{V}$ a $\sigma$-algebra?

Not if $A\cup U\notin\mathcal{A}$ and $A\cup U\notin\mathcal{V}$.

$\endgroup$

You must log in to answer this question.

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