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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.

A.Magnus
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    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." – High GPA Aug 17 '23 at 07:49

3 Answers3

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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

mm-aops
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  • 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? – A.Magnus Jan 27 '15 at 18:15
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    @A.Magnus ${a}\cup{b}={a,b}$ is not there. – drhab Jan 27 '15 at 18:21
  • @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? – A.Magnus Jan 27 '15 at 18:30
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    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 $\bigcup_{n=1}^{\infty}A_n={a,b}\notin\mathcal A$. – drhab Jan 27 '15 at 18:49
  • @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. – A.Magnus Jan 27 '15 at 19:31
  • If you like you can write $A_1=\emptyset$, $A_2={a}$ et cetera and come to the conclusion that $\bigcup_{n=1}^6 A_n=X$ (nothing against it), but what has that to do with checking whether we are dealing here with a sigma-algebra? Nothing! For this you must check what I mentioned in my former comment (and also it must be checked to be closed under complements, wich is the case here). – drhab Jan 27 '15 at 19:40
  • I think I misunderstood from the beginning the textbook's 4th. condition in $\sigma$-algebra: "Whenever $A_1, A_2, \ldots $ are in $\mathscr A$, then $\bigcup {i=1}^{\infty} A_i$ and $\bigcap _{i=1}^{\infty} A_i$ are in $\mathscr A$." So here $\infty$ should be read as "any number." I said this because in another posting I saw that this condiution should be read "...for any sequence $(X_i){i \in \mathbb ℕ}$ of elements in a $\sigma$-algebra, the union $⋃_{i \in \mathbb ℕ}X_i$ is in $\mathscr A$." Sorry for this long debate! Thanks! – A.Magnus Jan 27 '15 at 20:10
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$\{\emptyset,[0,3/4),[3/4,1),[0,1)\}$ 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.

zoli
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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 $A\cup U\notin\mathcal{V}$.

drhab
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