# Need help with Sigma-algebra

I am confused on how to determine a Sigma-algebra.

The following partitions of a set are given:
$$A1 = \{1,3\}$$ $$A2 = \{2,4,6,8\}$$ $$A3 = \{5,7,9\}$$ And Omega is $$\Omega = \{1,2,3,4,5,6,7,8,9\}$$

I know that at least Omega and the empty set has to be part of the Sigma-algebra as well as the single partitions and their complements: $$\sigma(A1,A2,A3) = \{\emptyset,\Omega,\{1,3\},\{2,4,5,6,7,8,9\},\{2,4,6,8\},\{1,3,5,7,9\},\{5,7,9\},\{1,2,3,4,6,8\}\}$$

Additionally, there has to be the union of the elements inside and this is the point I am not sure about. Are the following elements part of the Sigma-algebra as well? $$A1 \cup A2 = \{1,2,3,4,6,8\}$$ $$A2 \cup A3 = \{2,4,5,6,7,8,9\}$$ $$A1 \cup A3 = \{1,3,5,7,9\}$$ $$A1 \cup A2 \cup A3 = \Omega$$

I think I can ignore all of them since they are already in my Sigma-algebra.

Is my Sigma-algebra fullfinished and where my thoughts about creating the algebra correct? I know that there have to be 2^n elements inside the algebra and due to the fact of there are 8 elements this might be a correct solution.

Another question:

In case a union (e.g. A1 U A2) would not have been in the algebra. Do I just have to add A1 U A2 to the algebra or also its complement?

You're right on target. The ($\sigma$-)algebra on $\Omega$ generated by the partition $\{A_1,A_2,A_3\}$ is precisely $$\{\emptyset,A_1,A_2,A_3,A_1\cup A_2,A_1\cup A_3,A_2\cup A_3,\Omega\}.$$ More generally, if you're given a partition of any set $\Omega$ into $n$ subsets, the ($\sigma$-)algebra on $\Omega$ generated by that partition will consist of $2^n$ subsets, consisting of all finite unions of the partition elements (including the empty union).
• Thanks! Don't I have to unite all the partitions as well? $$A1 \cup A2 \cup A3$$ In my script there's something like $$\sigma(A1,...An) = \{A1^{\epsilon1} \cup A2^{\epsilon2} \cup ... \cup An^{\epsilon_n} : \epsilon_i \in \{0,1\}, i=1,...,n\}$$ and this has confused me. I thought I have to unite all elements as well. Commented Oct 26, 2013 at 13:16
• Yes, but $A_1\cup A_2\cup A_3=\Omega,$ so that's in there. As for the other thing, I suspect it is intended that $A^0=\emptyset$ and $A^1=A$, so (for example) $A_1\cup A_2=A_1^1\cup A_2^1\cup A_3^0$ has the desired form, as does $\emptyset=A_1^0\cup A_2^0\cup A_3^0,$ as does $\Omega=A_1^1\cup A_2^1\cup A_3^1.$ Commented Oct 26, 2013 at 13:23