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Let $\mathcal{A}$ and $\mathcal{B}$ be two countably generated $\sigma$-algebras on the set $X$. Is the $\sigma$-algebra $\mathcal{A} \cap \mathcal{B}$ necessarily countably generated?

I suspect that the answer is negative since all attempts to prove it failed and there seems to be no apparent reason for the statement to be true. On the other hand, I was unable to come up with a counterexample.

If it helps to obtain a positive answer, I don't mind assuming that singletons are measurable.


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up vote 6 down vote accepted

The wonderful booklet Borel Spaces by Rao and Rao gives a number of examples. Here is one of them:

Lemma: There is an injective function $f:[0,1]\to [0,1]$ such that the preimage of every uncountable Borel set with uncountable complement is not a Borel set.

Proof: There are continuum many Borel sets, so let $(A_0,B_0), (A_1,B_1),\ldots, (A_\alpha,B_\alpha),\ldots,$ with $\alpha<\mathfrak{c}$ be a well-ordering of all disjoint pairs of uncountable Borel sets. Constuct recursivelya $\mathfrak{c}$-sequence of pairs by choosing for each $\beta<\mathfrak{c}$ two points $x_\beta\in A_\beta$ and $y_\beta\in B_\beta$ that are not in $\{x_\alpha, y_\alpha;\alpha<\beta\}$ (this is possible since every uncountable Borel set has the cardinality of the continuum). Let $f(x_\alpha)=y_\alpha$, $f(y_\alpha)=x_\alpha$ and let $f(z)=z$ for all $z$ not occuring in the sequence. We now verify that $f$ has the required property. It is clear that $f$ is injective. Let $A$ be an uncountable Borel set with uncountable complement. Since there are uncountably many pairs of disjoint uncountable Borel subsets of $A$, we have that $D=f^{-1}(A)\cap A$ is uncountable. It has an uncountable complement, because $A$ has one. Suppose that $f^{-1}(A)$ is a Borel set. Then $D$ is an uncountable Borel set with uncountable complement, such that $D=f^{-1}(D)$. But this is impossible. The pair $(D,D^C)$ must show up in our well-ordering, so $f$ maps a point in $D$ to a point in $D^C$. $\blacksquare$

Corollary: Let $f$ be an injective function $f:[0,1]\to [0,1]$ such that the preimage of every uncountable Borel set with uncountable complement is not a Borel set. Let $\mathcal{D}$ be the $\sigma$-algebra generated by $f$ and $\mathcal{B}$ be the Borel $\sigma$-algebra. Then $\mathcal{D}\cap\mathcal{B}$ is the $\sigma$-algebra consisting of countable sets and sets with countable complements.

Proof: Every, set in $\mathcal{D}$ that is uncountable and has an uncountable complement must be the preimage of an uncountable Borel set with uncountable complement. But such preimages are by the last lemma never Borel sets. So every set in $\mathcal{D}\cap\mathcal{B}$ is countable or has a countable complement. Moreover, it is easy to see that every such set is in the intersection. $\blacksquare$

We show that the $\sigma$-algebra on $[0,1]$ consisting of countable sets and sets with countable complements is not countably generated (Rao & Rao prove this in a different way). First, we need a Lemma:

Lemma: Let $\mathcal{C}$ be a countably generated $\sigma$-algebra on a set $X$. Then there is a function $g:X\to [0,1]$ such that $\mathcal{C}$ is the $\sigma$-algebra generated by $g$ (where the codomain is endowed with the Borel $\sigma$-algebra). Moreover, if for all distinct $x,y\in X$ there is a $C\in\mathcal{C}$ such that $x\in C$ and $y\notin C$, the $g$ must be injective.

Proof: Let $\{C_0,C_1,C_2,\ldots\}$ be a countable set of generators. Let $g$ be given by $$g(x)=\sum_{n=0}^\infty\frac{2}{3^{n+1}}I_{C_n},$$ where $I_{C_n}$ is the indicator function of $C_n$. The function $g$ is known as the Marczewski-function. If $g(x)=g(y)$, then there is no nonempty set $B$ such that $x\in g^{-1}(B)$ but $y\notin g^{-1}(B)$. $\blacksquare$

Corollary: The $\sigma$-algebra on $[0,1]$ consisting of countable sets and those with countable complement is not countably generated.

Proof: Suppose it is. Then by the preceding lemma, the $\sigma$-algebra is generated by some $g:[0,1]\to[0,1]$. Let $x,y\in[0,1]$ be two different points. Then $x\in [0,1]\backslash\{y\}$ but $y\notin[0,1]\backslash\{y\}$. It follows that $g$ must be injective. Since $g$ is measurable, either $g^{-1}\big([0,1/2]\big)$ or $g^{-1}\big([1/2,1]\big)$ must have a countable complement. Say, it is is the latter. Again $g^{-1}\big([1/2,3/4]\big)$ or $g^{-1}\big([3/4,1]\big)$ must have countable complement. Continuing this way, we get a decreasing sequence of sets with countable complement. It follows that the intersection $I$ has a countable complement. But $I$ is the preimage of a singleton under $g$. This contradiction proves that the $\sigma$-algebra is not countably generated. $\blacksquare$

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Awesome, thank you so much! The proof of the first lemma is a beautiful variant of the Bernstein construction. With your second lemma (which I only knew in the injective case) I think I can carry out my idea using ergodicity. Let me check the arguments in a bit more detail before accepting your answer. – Mirjam Jan 21 '13 at 15:39
Marczewski introduced the Marczewski function in the 1936 article "Sur l'équivalence des suites d'ensembles et l'équivalence des fonctions" (link) (see section 1.2; the function is denoted $f_e$), where he also proves its measurability w.r.t. $\sigma\left(C_i\right)$ (see section 1.3). The article is signed by E. Szpilrajn. This was Marczewski's original, Jewish family name, which he changed after the Germans took over his homeland of Poland. – Evan Aad Apr 13 '13 at 20:12
By the way, in said article Marczewski credits his friend and mentor M. Kuratowski with obtaining the analytic formula of the Marczewski function (footnote on page 306), which should therefore really be called the Szpilrajn-Kuratowski or the Szpilrajn/Marczewski-Kuratowski function. – Evan Aad Apr 13 '13 at 20:32
@EvanAad Great info. I always wondered where the name change came from. – Michael Greinecker Apr 13 '13 at 21:34

I hope it's OK to provide the (negative) answer myself. I continued searching for an answer and found:

Remarks on some Borel structures
Grzegorek, E.
Measure Theory Oberwolfach 1983
Lecture Notes in Mathematics Volume 1089, 1984, pp 69-74

In the first section that paper discusses variants of results (under the continuum hypothesis) by B. V. Rao that imply a negative answer to my question.

Rao, B.V.
On Borel structures. (English)
Colloq. Math. 21, 199-204 (1970).

The original article by B. V. Rao does not appear to be available online, but here's the author's summary from the Zentralblatt:

The purpose of this paper is to record certain observations about Borel structures. After giving the relevant definitions, we show that every countably generated Borel space has a minimal generator, that the intersection of two separable Borel structures need not be separable, and that there are atomless Borel structures on any uncountable set. Fionally, we give some applications of a theorem of D. Blackwell [Proc. 3rd Berkeley Sympos. math. Statist. Probab. 2, 1-6 (1956, this Zbl. 73, 123)] and G. W. Mackey, [Trans. Amer. math. Soc. 85, 134-165 (1957, this Zbl. 112)], after a precise statement of the theorem.

A $\sigma$-algebra is separable if it is countably generated and contains the singletons.

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If you want, I can post a proof tomorrow. – Michael Greinecker Jan 21 '13 at 0:10
Thank you, that would be fantastic! After I found out what to look for, I found a hint in Bogachev's book (exercise 6.10.37 on page 58 of vol. II). Apparently one can take the $\sigma$-algebras of Borel sets in $\mathbb{R}$ invariant under translations by $1$ and $\pi$, respectively. I haven't managed to figure out the details, yet. Unfortunately, our local library doesn't have the journal Colloq. Math. and neither does it have the book by Rao and Rao, Bogachev refers to. – Mirjam Jan 21 '13 at 0:32
I think I managed to figure out how to use ergodicity of translation by $\pi$ on $\mathbb{R/Z}$ to show that the intersection of the two $\sigma$-algebras does not admit a countable generating set that also separates points. However, so far, I was unable to show that there is no countable generating set in general. I suppose I miss something really simple. – Mirjam Jan 21 '13 at 13:27
The booklet by Rao & Rao doesn't provide any explanation either, it says no more then the hint given by Bogachev. – Michael Greinecker Jan 21 '13 at 15:17

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