Let $(X,\Sigma)$ be a measurable space of arbitrary cardinality. I would like to understand under which conditions this space is isomorphic to a measurable subset of $\{0,1\}^\kappa$ for some cardinal $\kappa$.

To clarify, by "isomorphic to a measurable subset", I mean that there should exist a measurable injection $X\hookrightarrow\{0,1\}^\kappa$ which in addition takes measurable sets to measurable sets. As pointed out in the comments, there are two $\sigma$-algebras which one can put on $\{0,1\}^\kappa$, and these two differ for uncountable $\kappa$:

  1. The Baire $\sigma$-algebra, which is the smallest $\sigma$-algebra making the product projections $\{0,1\}^\kappa\to\{0,1\}$ measurable;
  2. The Borel $\sigma$-algebra, which is the one generated by the product topology.

These two give rise to different versions of my question, and it would be optimal to have an answer in both cases, although I currently find the Baire $\sigma$-algebra more natural.

For example, if $(X,\Sigma)$ is the standard Borel space, then either kind of embedding exists since $\{0,1\}^{\mathbb{N}}$ is itself a standard Borel space.

Clearly, a necessary condition for either embedding to exist is that $(X,\Sigma)$ needs to be separated in the sense that for all $x,y\in X$, there is $A\in\Sigma$ with $x\in A$ and $y\not\in A$.

Is separatedness sufficient for $(X,\Sigma)$ to embed into some $\{0,1\}^\kappa$, equipped with either the Baire or the Borel $\sigma$-algebra?

Motivation: I am working with Bayesian networks containing latent variables. I want to allow the sample space of a latent variable to be an arbitrary probability space. Since the latter are difficult to work with, I am trying to make an argument along the lines of replacing every latent variable by a collection of binary variables. A positive answer to the above question would be one way to do this.

  • $\begingroup$ Could you clarify the question a bit more precisely? When you say "a measurable subset of $\{0,1\}^\kappa$", do you mean a Borel subset of that space, considered with the product topology? And then you want the isomorphism to take the sets in $\Sigma$ exactly to the Borel subsets of that Borel set? $\endgroup$ Feb 27 '14 at 20:40
  • $\begingroup$ @Joel: yes, that's exactly what I mean. I'll clarify the question accordingly. $\endgroup$ Feb 27 '14 at 20:44
  • $\begingroup$ The smallest $\sigma$-algebra making each projection measurable is not the Borel $\sigma$-algebra when $\kappa$ is uncountable but rather the Baire $\sigma$-algebra. For instance, every one-point set is a Borel set, but not a Baire set. I am therefore confused about whether this question refers to the Borel sets of the Baire sets. $\endgroup$ Feb 27 '14 at 21:19
  • $\begingroup$ @Joseph: you're right. I will have to think about it a bit more. Sorry for the confusion... $\endgroup$ Feb 27 '14 at 21:31
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    $\begingroup$ I think the countable and cocountable sets would be a counterexample for the Borel sets. $\endgroup$ Feb 27 '14 at 22:02

Let $X$ be a set and give $X$ the discrete topology. Give $X\cup\{\infty\}$ the one-point compactification $X$. Then every subset $U$ of $X$ is open in $X\cup\{\infty\}$. Therefore, every subset of $U$ of $X\cup\{\infty\}$ is a Borel set. On the other hand, $X\cup\{\infty\}$ can be embedded as a closed subspace of $2^{\kappa}$ for some $\kappa$. Take note that the Borel subsets of $X\cup\{\infty\}$ are simply the restrictions of the Borel sets to $X\cup\{\infty\}$. Therefore, the power set $\sigma$-algebra $(X\cup\{\infty\},P(X\cup\{\infty\})$ can be embedded as Borel subset of $2^{\kappa}$.

A closer examination of my argument shows that a if $X$ can be embedded as a Borel set of some compactification (for example for locally compact spaces), then the $\sigma$-algebra $(X,Bor(X))$ (here $Bor(X)$ stands for the collection of all Borel subsets of $X$) can be embedded as a Borel subset of $2^{\kappa}$.

On the other hand, if $B\subseteq 2^{\kappa}$ is a Borel set, then the Borel measure on $B$ is the Borel measure of the topology on $B.$ However, by this answer not every $\sigma$-algebra is the Borel measure of some topology.


As in Joseph Van Name's comment, let $X$ be an uncountable set and $\Sigma$ the $\sigma$-algebra of all countable and cocountable sets. Then $(X,\Sigma)$ is separated.

First note that $(X,\Sigma)$ is not standard Borel (it isn't countably generated), so it does not embed into $\{0,1\}^{\aleph_0}$ with its Borel/Baire $\sigma$-algebra.

On the other hand, if $\kappa$ is uncountable, the Baire $\sigma$-algebra $\mathcal{B}_0$ on $\{0,1\}^\kappa$ consists of sets depending on at most countably many coordinates. In particular, every nontrivial set in $\mathcal{B}_0$ is uncountable and has uncountable complement. So no injection $X \to \{0,1\}^\kappa$ can map measurable sets to measurable sets.

I'm still thinking about the Borel $\sigma$-algebra.

  • $\begingroup$ Neat answer! I can't accept both, so I will go with Joseph's which covers the Borel case, since it also contains a sufficient condition which is close to his necessary one. $\endgroup$ Feb 28 '14 at 12:55

How do you like this construction:

Let $X$ be a set and $\Sigma$ a sigma-algebra on $X$ that separates points of $X$. For each $E \in \Sigma$, define $\phi_E \colon X \to \{0,1\}$ by $\phi_E(x) = 1$ if $x \in E$ and $\phi_E(x) = 1$ of $x \notin E$. (The characteristic function of $E$.) Then define $\Phi \colon X \to \{0,1\}^\Sigma$ by $$ \Phi(x)(E) = \phi_E(x)\qquad\text{for all } E \in \Sigma $$ Consider the product sigma-algebra $\Omega$ on $\{0,1\}^\Sigma$. Then investigate whether, for all $A \subseteq X$, $$ A \in \Sigma \quad\Longleftrightarrow\quad A = \Phi^{-1}(Q) \text{ for some } Q \in \Omega . $$

  • $\begingroup$ That's precisely what I had tried initially (see the original version of my question). If you know about Gelfand duality and Stone duality, it seems like a natural thing to try and indeed the resulting injective function is measurable (with respect to the Baire, and hence also the Borel $\sigma$-algebra on $\{0,1\}^\Sigma$). However, I don't see why this function should also take measurable sets to measurable sets, and I think this fails in general. In fact, I suspect that its image may not even be measurable, but I don't have a concrete example of that. $\endgroup$ Feb 28 '14 at 19:24

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