Given a measure of some subsets, how can you determine the number of sets in the sigma algebra?

Question

I was given a question that I can't wrap my head around.

Given a measure space ($X, \Omega,\mu$), with $X$ = {1,2,3, ... 16 }, we have

$C_1$ = {1,2,3,4,5,6,7,8 }

$C_2$ = {9,10,11,12,13,14,15,16}

$C_3$ = {1,2,5,6,9,10,13,14}

$C_4$ = {3,4,7,8,11,12,15,16}

$C$ denotes the field generated by {$C_1,C_2,C_3,C_4$}. We then let $\Omega$ = $\sigma[C]$.

With $\mu(C_\mathscr i) = 1/2, 1\le\mathscr i\le4$, and $\mu(C_1\cap C_3) = 1/4$, how can I show that $\hat{\Omega_\mu} = \Omega$, with $2^4 = 16$ sets?

I have tried equating the measures of the subsets to 1, but I do not know how it will determine the number of sets in $\Omega$. Additionally, I suppose that the proving of the completeness of $\Omega$ (hence it being equal to $\hat{\Omega_\mu}$) will come after I get the number of subsets in the sigma algebra? Please enlightment me. Thank you so much!

Answer 1

Let $\mathcal{C}$ be a family of subsets of some finite set $X$, and let $\Omega$ be the $\sigma$-algebra generated from $\mathcal{C}$. Let $\mathcal{A}$ be the set of subsets of $S \subseteq X$ that are "indivisible" in the sense that they always go together as a group, or not at all, in any $C \in \mathcal{C}$, and that are as big as possible. More formally, define $\mathcal{A}$ be the set of subsets of $S$ so that

1. For any set $C \in \mathcal{C}$, either $S \cap C = \emptyset$ or $S \cap C = C$.
2. There is no proper superset of $S$ that has property (1).

I claim that $|\Omega| = 2^{|\mathcal{A}|}$. I'll give an outline for the proof and leave the details to you.

The reason is that we have a bijection $\mathcal{P}(\mathcal{A}) \rightarrow \Omega$, where $\mathcal{P}(\mathcal{A})$ is the power set of $\mathcal{A}$, given by $f(T) = \bigcup_{A \in T} A$. To see this, we must show

1. $f(T) \in \Omega$ for any $T \in \mathcal{P}(\mathcal{A})$. To show this, first show that any $A \in \mathcal{A}$ is an intersection of elements of $C$; specifically, $A = \cap_{C \in \Omega, a \subseteq C} C$. That follows because of property (2) (prove this!). Then $f(T)$ is a union of intersections of elements of $C$ and so belongs to the $\sigma$-algebra.

2. $f$ is surjective. To see this, prove that the image of $f$ is a $\sigma$-algebra and contains $\mathcal{C}$. It follows from the definition of a generated $\sigma$-algebra that the image of $f$ contains $\Omega$. (do you see why?)

3. $f$ is injective. We must show that if $f(T_1) = f(T_2)$ then $T_1 = T_2$. Hint: Pick an element of $f(T_1)$, say $x$; then there must be some $A_1 \in T_1$ and $A_2 \in T_2$ so that $x \in A_1$, $x \in A_2$. Show $A_1 = A_2$ and continue, inductively.