# Finite $\sigma$-algebras are generated by unique minimal finite partitions

I am trying to prove the following statement:

Let $$(X,\mathcal A)$$ be a measure space s.t. the $$\sigma$$-algebra $$\mathcal A$$ is finite, then there exists a unique minimal finite partition $$\mathcal P=\{P_1,\dots,P_n\}$$ of $$X$$ s.t. $$\sigma(\mathcal P)=\mathcal A$$ and for all $$A\in\mathcal A$$ holds $$(*)\qquad A\cap P_k\in\{\emptyset,P_k\}.$$

Here's what I have so far:

Let $$\mathcal A=\{\emptyset,X,A_1,\dots,A_m\}$$, where $$A_i$$ are the nontrivial elements of the algebra (assume $$m>0$$, there is nothing to prove for $$\mathcal A=\{\emptyset, X\}$$), then $$\bigcup_iA_i=X$$ (otherwise $$\left(\bigcup_iA_i\right)^c=A_j$$ for some $$j$$, which is a contradiction). Disjointise the $$A_i$$ and get $$B_i\in\mathcal A$$ s.t. $$\bigcup_iB_i=X$$ and $$B_i\cap B_j=\emptyset$$ for $$i\neq j$$ (standard measure theoretic trick). Each $$B_i$$ is either empty or equal to $$A_{k_i}$$ for some $$k_i$$, so discard the empty ones and get $$P_j:=A_{k_j},1\leq j\leq n$$, which parition $$X$$. For $$A\in\mathcal P$$ the condition (*) is trivial [for $$A\in\mathcal A$$ arbitrary I'm having trouble showing this]. Now, assuming (*), we get for an arbitrary $$A\in\mathcal A$$: $$A=A\cap X=\bigcup_jA\cap P_j=\bigcup_kP_k,$$ so $$A\in\sigma(\mathcal P)$$ and therefore $$\mathcal A=\sigma(\mathcal P)$$.

I'm having trouble showing that the partition I get this way is both unique and minimal and that it has property (*). Any tips on how to proceed?

Also, if $$\mathcal A,\mathcal P$$ are as above, would it be correct to say that $$|\mathcal A|=2^{|\mathcal P|}$$? It seems plausible to me that intersection and complementation in this situation do not produce anything other than (disjoint) unions of elements of $$\mathcal P$$, so the number of elements of $$\mathcal A$$ should be given by the number of subsets of $$\mathcal P$$ (each subset $$\{P_{k_1},\dots,P_{k_n}\}$$ determines the element $$\bigcup_iP_{k_i}$$ of $$\mathcal A$$). In particular, the cardinality of a finite $$\sigma$$-algebra is always a power of $$2$$. Would this be a correct argument?

• Try to assign to each sequence $s$ in $\{0,1\}^m$ the element $\bigcap_{i=1}^m A_i^{(s_i)}$ where $A^{(0)}=A^\complement$ and $A^{(1)} = A$ for any subset $A$ of $X$. These are disjoint for distinct sequences... Commented Oct 21, 2021 at 21:05
• The second part is fine, $|\mathcal A|=2^{|\mathcal P|}$. For the first part, rather consider all the minimal nonempty elements of $\mathcal A$. Commented Oct 21, 2021 at 21:07
• @HennoBrandsma your approach yields a partition with the desired property, but I am yet unsure how to prove minimality and uniqueness. Commented Oct 21, 2021 at 21:30
• minimality is clear (assuming we enumerated all non-unit elements). Unicity less so. Commented Oct 21, 2021 at 21:31

Since $$\mathcal A$$ is finite, it has minimal nonempty elements $$P_i$$ (with respect to inclusion): if a nonempty set in $$\mathcal A$$ is not minimal, it contains strictly smaller element of $$\mathcal A$$, and so on, but because $$\mathcal A$$ is finite, this procedure must stop (after at most $$|\mathcal A|-1$$ steps).
Observe that for any $$x\in X$$ there's a (unique) minimal element $$P_i$$ of $$\mathcal A$$ that contains $$x$$: in the above procedure, starting with $$X$$, because of closedness under set difference, we can always choose a smaller element which still contains $$x$$.
Or, put another way, $$\bigcap\{A\in\mathcal A:x\in A\}$$ is minimal.
Consequently, for any $$A\in\mathcal A$$ we have $$A=\bigcup\{P_i:P_i\subseteq A\}\,.$$