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?

  • 1
    $\begingroup$ 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... $\endgroup$ Commented Oct 21, 2021 at 21:05
  • 1
    $\begingroup$ The second part is fine, $|\mathcal A|=2^{|\mathcal P|}$. For the first part, rather consider all the minimal nonempty elements of $\mathcal A$. $\endgroup$
    – Berci
    Commented Oct 21, 2021 at 21:07
  • $\begingroup$ @HennoBrandsma your approach yields a partition with the desired property, but I am yet unsure how to prove minimality and uniqueness. $\endgroup$ Commented Oct 21, 2021 at 21:30
  • $\begingroup$ minimality is clear (assuming we enumerated all non-unit elements). Unicity less so. $\endgroup$ Commented Oct 21, 2021 at 21:31

1 Answer 1


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 $$A=\bigcup\{P_i:P_i\subseteq A\}\,.$$


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