# Approximating an arbitrary $\sigma$-algebra by simpler $\sigma$-algebras

A $\sigma$-algebra $\mathcal F$ over $\Omega$ is generated by an countable partition if there exits a countable partition $\mathcal B = \{ B_i \}$ of $\Omega$ such that $\mathcal F = \sigma(\mathcal B)$. Now let $\mathcal G$ be an arbitrary $\sigma$-algebra over $\Omega$. Is it possible to find $\sigma$-algebras $\mathcal G_n$ generated by countable partitions approximating $\mathcal G$, i.e. such that $G_1 \subseteq G_2 \subseteq G_3 \subseteq \ldots$ and $\mathcal G = \bigcup_n \mathcal G_n$?

One naive idea of me is to take some arbitrary $A_1 \in \mathcal G$, and then set $\mathcal B_1 = \{ A_1, A_1^C \}$ and $\mathcal G_1 = \sigma(\mathcal B_1)$, then take some $A_2 \in \mathcal G$ with $A_2 \cap A_1 = \emptyset$ and set $\mathcal B_2 = \{ A_1, A_2, (A_1\cup A_2)^C \}$ and $\mathcal G_2 := \sigma(\mathcal B_2)$ and so on. But this will not work, for example if $\mathcal P(\mathbb N) \subseteq \mathcal G$ and I choose $A_i := \{ i \}$, then as the "limit" $\sigma$-algebra I will get $\sigma(\{ A_i \}) = \{ M : M \subseteq \mathbb N \mbox{ or } X\setminus M \subseteq \mathbb N \}$, which has not to be the the original $\sigma$-algebra with which I started, for example if $\mathbb G = \mathcal B(\mathbb R)$ if fullfills $\mathcal P(\mathbb N) \subseteq \mathcal G$.

• I am very confused by your question. Is the property that you want satisfied by the $\sigma$-algebra of Borel subsets of $\mathbb R$? Is your property satisfied by the $\sigma$-algebra of Lebesgue-measurable subsets of $\mathbb R$? – André Henriques Jan 23 '15 at 14:53

## 4 Answers

It turns out that it is impossible to represent any $\sigma$-complete Boolean algebra as a countable union of Boolean subalgebras.

For example, in the paper, Boolean algebras as unions of chains of subalgebras by Sabine Koppelberg, the following result is proven.

$\mathbf{Theorem}$ Suppose $B$ is a Boolean algebra such that whenever $R,S\subseteq B$ are countable subsets and $r\wedge s=0$ for $r\in R,s\in S$ then there is some $b\in B$ where $r\leq b$ for each $r\in R$ and $s\leq b'$ for each $s\in S$. Then whenever $B_{n}$ is an increasing sequence of proper subalgebras of $B$, then $B\neq\bigcup_{n}B_{n}$.

Therefore to represent a $\sigma$-algebra as a sequence of simpler Boolean algebras, one would need a directed system of Boolean subalgebras or at least an uncountable sequence of partitions.

I also must mention that approximating Boolean algebras by partitions has been a fruitful endeavor for me. My dissertation (Boolean Partition Algebras 2013) and some of my further research has dealt with representing Boolean algebras as directed unions of subalgebras which are generated by partitions. Define a Boolean partition algebra to be a pair $(B,F)$ such that $B$ is a Boolean algebra and $F$ is a filter on the meet-semilattice of partitions of $B$ such that $B=\{0\}\cup\bigcup F$. If $p\in F$, then define $p^{*}=\{\bigvee R|R\subseteq p,\bigvee R\,\,\textrm{exists}\}$. It is not to hard to show that each $p^{*}$ is a Boolean subalgebra of $B$. Furthermore, it is easy to show that if $(B,F)$ is a Boolean partition algebra, then $B=\bigcup_{p\in F}p^{*}$. We say that a partition $p$ is subcomplete if the Boolean algebra $p^{*}$ is a complete Boolean algebra. I found Boolean partition algebras to be useful since the category of Boolean partition algebras is equivalent to other categories such as the category of all complete non-Archimedean uniform frames and it is contravariantly equivalent to the category of pro-sets with surjective transitional mappings (pro-sets are inverse systems of sets). Furthermore, ideas such as taking quotient Boolean algebras and taking subalgebras translate nicely into taking sub-inverse systems and taking quotient inverse systems.

Furthermore, the notion of a $\sigma$-ideal, $\sigma$-subalgebra, and $\sigma$-homomorphism between $\sigma$-complete Boolean algebras can all be represented as special ideals, subalgebras, and homomorphisms between Boolean partition algebras.

• Thanks for all answers! Is there any other way or notion in which general $\sigma$-algebras could be approximated by certain simpler $\sigma$-algebras? – StefanH Jan 23 '15 at 20:29
• Stefan. Besides representing Boolean algebras in terms of direct limits of power set algebras, there are nice ways to represent Boolean algebras in terms of products. A partition $p$ of a Boolean algebra $B$ is said to be a complete partition if whenever $c_{a}\leq a$ for each $a\in p$ the least upper bound $\bigvee_{a\in p}c_{a}$ exists. The complete partitions are precisely the partitions where the mapping $B\rightarrow\prod_{a\in p}B\upharpoonright a$ is an isomorphism of Boolean algebras. In fact, any decomposition of a Boolean algebra into a product is a decomposition of this form. – Joseph Van Name Jan 23 '15 at 23:14
• For example, every partition of a complete Boolean algebra is a complete partition. Furthermore, under the refinement ordering, the complete partitions are closed under finite meets they are upwards closed. In other words, the complete partitions form a filter on the meet-semilattice of partitions of a Boolean algebra. – Joseph Van Name Jan 23 '15 at 23:23

$\def\cG{\mathcal G}$The algebra $\cG=\mathrm{Borel}(\mathbb R)$ does not have this property.

Assume for contradiction that $\cG=\bigcup_n\cG_n$, where $\cG_0\subseteq\cG_1\subseteq\cG_2\subseteq\cdots$, and each $\cG$ is generated by a partition $\mathcal B^n=\{B^n_i:i\in\mathbb N\}$ of $\mathbb R$. Note that $\cG_n=\bigl\{\bigcup_{i\in I}B^n_i:I\subseteq\mathbb N\bigr\}$. For every $k\in\mathbb N$, we can find $n_k\ge k$ such that $(k,k+1)\in\cG_{n_k}$, hence there is $i_k$ such that $B^{n_k}_{i_k}$ is an infinite subset of $(k,k+1)$. Fix a nonempty Borel proper subset $C_k\subset B^{n_k}_{i_k}$, and put $C=\bigcup_kC_k$. Then $C$ is again Borel, but by construction, it is not in any $\cG_{n_k}$; since the $n_k$ are cofinal in $\mathbb N$, it is not in any $\cG_n$.

Interesting Boolean algebras tend to be not expressible as countable unions of subalgebras. Here's a Banach-space condition proved by Schachermayer - see Proposition 4.6 in

W. Schachermayer, On some classical measure-theoretic theorems for nonsigma-complete Boolean algebras, Dissertationes Math. 214 (1982).

which says that if $A$ is a Boolean algebra such that $A=\bigcup_{n=1}^\infty A_n$ for some chain of proper subalgebras $(A_n)_{n=1}^\infty$, then $C({\rm Stone}\, A)$ is not a Grothendieck space.

It is not totally elementary but also not very difficult to prove that Stone spaces of $\sigma$-algebras are $F$-spaces and for such spaces the space of continuous functions is Grothendieck. Of course in the case of $\sigma$-algebras direct proofs of such non-expressability exist:

A. Broughton, B. Huff, A comment on unions of sigma fields, American Mathematical Monthly 84, 7, (1977), 553-554.

The $\sigma$-algebra $\mathcal G$ does not need not be countably generated, so the answer is, obviously, "no".

On the other hand, if $\mathcal G$ is generated by subsets $E_n$, you can easily construct the $\mathcal G_n$ so that for each $n$, $\mathcal G_n$ is in the $\sigma$ algebra generated by $E_n$ and $\mathcal G_{n-1}$ (use the partition $\{A\cap E_n, A\cap E_n^c : A \in \mathcal P_{n-1}\}$, where $P_n$ is a partition that generates $\mathcal G_{n-1}$).

• "$\mathcal G$ need not be countably generated" - that is not said, or is every countable union of countable generated $\sigma$-algebras countable generated? – StefanH Jan 23 '15 at 14:44
• I may be confused, but even if $\mathcal G$ is generated by $(E_n)_{n\in\mathbb N}$, it is not necessarily the case that every $E\in\mathcal G$ is in the $\sigma$-algebra generated by $(E_n)_{n\le m}$ for some $m$. Consider for example the Borel algebra on $\mathbb R$, generated by intervals with rational end-points. – Emil Jeřábek Jan 23 '15 at 14:44
• Maybe I misinterpreted the question. I thought you wanted $\mathcal G$ to be generated by the union of the $\mathcal G_n$. If you want it actually to be the union of the $\mathcal G_n$, then of course it is impossible, as Joseph and Emil pointed out. – Bill Johnson Jan 23 '15 at 15:25