# Are $\sigma$-algebras that aren't countably generated always sub-algebras of countably generated $\sigma$-algebras?

It is well known that there are countably generated sigma-algebras containing sub-sigma-algebras that cannot be countably generated. (Some tail sigma-algebras serve as examples.)

Question. Suppose $\mathcal{A}$ is a sigma-algebra of subsets of $X$ that cannot be countably generated. Does there exist a countably generated sigma-algebra $\mathcal{B}$ on $X$ such that $\mathcal{A}$ is a sub-$\sigma$-algebra of $\mathcal{B}$?

I'm inclined to think, but don't know how to prove, that the answer is no. If that's right, under what conditions can the desired $\mathcal{B}$ be found?

The answer is no as there is a cardinality bound for countably generated $\sigma$-algebras (it is continuum). You may then pick any $\sigma$-algebra of greater cardinality for a counter-example.

To see that it is continuum indeed note that every countably generated $\sigma$-algebra is of the form $$\{f^{-1}(B)\colon B\in \text{Borel}([0,1])\}$$ for some $[0,1]$-valued measurable function $f$ on your measurable space.

• Isn't $\aleph_1$ another bound? – nombre Dec 8 '16 at 13:07
• @nombre, not sure what do you mean by that. Every infinite $\sigma$-algebra has cardinality at least continuum. – Tomek Kania Dec 8 '16 at 13:11
• Oh yes you're right, I was thinking of the construction of the generated $\sigma$-algebra from the bottom up as a union indexed by $\omega_1$ but I didn't realize that at each step you may add ${\kappa}^{\aleph_0}$ measurable sets where $\kappa$ is the cardinal of the previous step. – nombre Dec 8 '16 at 13:16
• Do you know if there is a family $\mathcal A\subset\mathcal P(\mathbb R)$ such that $|\mathcal A|\le2^{\aleph_0}$ and $\mathcal A$ is not contained in any countably generated $\sigma$-algebra? Is there a countably generated $\sigma$-algebra which contains all the analytic subsets of $\mathbb R?$ – bof Dec 29 '16 at 6:56

Notation: (i). $P(X)$ is the set of all subsets of $X.$ (ii). $[y]^{\leq \omega}$ is the set of all countable (finite or countably infinite ) subsets of $y.$ (iii). For $S\subset P(X)$ let $B_X(S)$ be the Boolean algebra on $X$ generated by $S.$ (iv). $|t|$ denotes the cardinal of the set $t.$

Take $X$ such that $|P(X)|> 2^{\omega}.$ Now $P(X)$ is the largest $\sigma$-algebra on $X.$ And $P(X)$ is not countably generated.

Proof: Take any $T\in [P(X)]^{\leq \omega}.$ Let $T(0)=B_X(T).$ For ordinal $a<\omega_1$ let $$T(a+1)=B_X(\;\{\cup S: S\in [T(a)]^{\leq \omega}\}\;).$$ For $0<a=\cup a< \omega_1$ let $$T(a)=B_X(\;\cup_{b<a}T(b)\;).$$ By transfinite induction (and using $(2^{\omega})^{\omega}=2^{\omega}$) we have $$\forall a<\omega_1\;(|T(a)|\leq 2^{\omega}).$$ Let $U_T=\cup_{a<\omega_1}T(a).$ Then $U_T$ is a $\sigma$-algebra on $X.$

For if $Y\in [U_T]^{\leq \omega},$ then for each $y\in Y$ let $f(y)$ be the least (or any) $a\in \omega_1$ such that $y\in T(a).$ Since $\omega_1$ is regular and $Y$ is countable, there exists $b< \omega_1$ with $0<b=\cup b,$ such that $\forall y\in Y\; (f(y)<b).$ So $$\{T(f(y)):y\in Y\}\in [\cup_{c<b}T(c)]^{\leq \omega}$$ $$\text {which implies }\quad \cup Y\in T(b+1)\subset U_T.$$

Since $U_T$ is a sigma-algebra on $X$ and $T\subset U_T$ we have $$B_X(T)\subset U_T.$$

Now $|U_T|\leq |\cup_{a<\omega_1}(\{a\}\times T(a))|\leq$ $\omega_1\cdot 2^{\omega}=$ $2^{\omega}<|P(X)|.$ So $B_X(T)\subset U_T\ne P(X).$

Here is an example of a $\sigma$-algebra of cardinality $2^{\aleph_0}$ which is not a subalgebra of a countably generated $\sigma$-algebra. (The previously posted examples had cardinality greater than $2^{\aleph_0}.$)

Consider the space $X=\{0,1\}^\mathbb R$ with the Tychonoff product topology, where $\{0,1\}$ has the discrete topology. Let $\mathcal A$ be the $\sigma$-algebra generated by the clopen subsets of $X.$

To see that $|\mathcal A|=2^{\aleph_0},$ we start by noting that (since $X$ is compact) a clopen set in $X$ is a finite union of basic open sets, and there are just $2^{\aleph_0}$ of those. Now the argument showing that a countably generated $\sigma$-algebra has cardinality at most $2^{\aleph_0}$ also shows that a $\sigma$-algebra generated by $2^{\aleph_0}$ sets has cardinality $2^{\aleph_0}.$

I claim that $\mathcal A$ is not contained in any countably generated $\sigma$-algebra of subsets of $X.$ Suppose $\mathcal B$ is the $\sigma$-algebra generated by a countable family $\mathcal D\subset\mathcal P(X);$ I have to show that $\mathcal A\not\subseteq\mathcal B.$ Define a mapping $f:X\to\mathcal P(D)$ by setting $f(x)=\{D\in\mathcal D:x\in D\}.$The mapping $f$ is not injective, since $|X|=2^{2^{\aleph_0}}\gt2^{\aleph_0}=|\mathcal P(D)|.$ Thus there are points $x,y\in X$ such that $x\ne y$ but $f(x)=f(y).$ Then $x$ and $y$ are not separated by any set in $\mathcal D,$ nor by any set in $\mathcal B,$ the $\sigma$-algebra generated by $\mathcal D.$ On the other hand, there is a clopen set of $X$ which separates $x$ from $y;$ that clopen set belongs to $\mathcal A$ but not to $\mathcal B.$