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If $\left(\Omega_0,\mathcal{A}_0\right)$ is a Borel space, the set $B:=\left\{\left(\omega,\omega\right)\space:\mid\space\omega\in\Omega_0\right\}$ is measurable in $\mathcal{A}_0\otimes\mathcal{A}_0$. Equivalently, if $f,g:\left(\Omega_1,\mathcal{A}_1\right)\rightarrow\left(\Omega_0,\mathcal{A}_0\right)$ are measurable, the set $\left\{f=g\right\}$ is measurable in $\mathcal{A}_1$. But this need not hold if $\left(\Omega_0,\mathcal{A}_0\right)$ is not a Borel space.

Question #1 What conditions on $\left(\Omega_0,\mathcal{A}_0\right)$ guarantee the measurability of $B$? In particular, if $\mathcal{A}_0$ contains all the singletons of $\Omega_0$, will $B$ be measurable necessarily?

More generally, it can be shown (e.g. by using projections with the equivalent formulation indicated above) that if $\left(\Omega_i,\mathcal{A}_i\right)$ are Borel spaces for $i\in\mathbb{N}$, then for all $n\in\mathbb{N}$, the sets $\left\{\left(\omega,\omega\right)\space:\mid\space\omega=\left(\omega_1,\dots,\omega_n\right)\in\times_{i=1}^n\Omega_i\right\}$ are measurable in $\mathcal{A}^{(n)}\otimes\mathcal{A}^{(n)}$ with $\mathcal{A}^{(n)}:=\otimes_{i=1}^n\mathcal{A}_i$ and the set $\left\{\left(\omega,\omega\right)\space:\mid\space\omega\in\times_{i=1}^\infty\Omega_i\right\}$ is measurable in $\mathcal{A}^{\left(\infty\right)}\otimes\mathcal{A}^{\left(\infty\right)}$ with $\mathcal{A}^{\left(\infty\right)}:=\otimes_{i=1}^\infty\mathcal{A}_i$.

Question #2 Does this result extend to the case when $\mathbb{N}$ is substituted by an arbitrary uncountable index set $\mathcal{I}$ (but the $\left(\Omega_i,\mathcal{A}_i\right)$ are still Borel spaces)?

Question #3 Does the equivalence of the two formulations mentioned in the first paragraph still hold when the index set is uncountable? In other words, is it possible for $B=\left\{\left(\omega,\omega\right)\space:\mid\space\omega\in\times_{i\in\mathcal{I}}\Omega_i\right\}$ to be non-$\mathcal{A}^{\left(\mathcal{I}\right)}\otimes\mathcal{A}^{\left(\mathcal{I}\right)}$-measurable (with $\mathcal{A}^{\left(\mathcal{I}\right)}:=\otimes_{i\in\mathcal{I}}\mathcal{A}_i$) even as the set $\left\{f=g\right\}$ is measurable for all measurable $f,g:\left(\Omega,\mathcal{A}\right)\rightarrow\left(\times_{i\in\mathcal{I}}\Omega_i,\otimes_{i\in\mathcal{I}}\mathcal{A}_i\right)$?

Evan Aad
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  • What do you mean by a Borel space? Mackey used the term for measurable spaces. – Michael Greinecker Apr 13 '13 at 08:18
  • @MichaelGreinecker: A measurable space $S=(\Omega,\mathcal{A})$ is called a Borel space iff there exists a Borel set $C$ on the real line such that $S$ and $\left(C,\mathfrak{B}(C)\right)$ are isomorphic (isomorphism being a bi-measurable bijection; $\mathfrak{B}(C)$ is the restriction of the Borel field on the real line to subsets of $C$). This is the way it is defined in, say, Klenke (Definition 8.34), Kallenberg (just before Theorem A1.6) and Schervish (Definition B.31). – Evan Aad Apr 13 '13 at 08:39

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That the diagonal in $\Omega\times\Omega$ is measurable is equivalent to the identity function on $\Omega$ having a measurable graph. Take a measurable function $f:S\to T$ with $(S,\mathcal{S})$ and $(T,\mathcal{T})$ being measurable spaces. A necessary and sufficient condition for $f$'s graph to be measurable is that there exists a countably generated sub-$\sigma$-algebra $\mathcal{C}\subseteq\mathcal{T}$, such that $\mathcal{C}$ contains all singletons from $f(S)$. For a proof, see Proposition 2.1 here. So your set $B$ is measurable if and only if there is a countably generated sub-$\sigma$-algebra that contains all singletons.

A necessary condition for this is that $\Omega$ has cardinality no more than the continuum. To see this, use the fact that every countably generated $\sigma$-algebra is generated by a real-valued random variable. I have reported a proof here. If all $\Omega_i$ contain more than one element, if we take the product of at least continuum many of them, their uncountable product is automatically larger than the continuum.

Measurability of singletons will also be a problem with uncountable products. Take the set $\{0,1\}$ endowed with the discrete $\sigma$-algebra. Let $\kappa$ be uncountable and endow $\{0,1\}^\kappa$ with the product $\sigma$-algebra. Then no singleton is measurable. This follows from the fact that every set in the product-$\sigma$-algebra is generated by countably many coordinates. This again follows from the general result that if $A\in\sigma(\mathcal{F})$ then there exists a countable family $\mathcal{D}\subseteq\mathcal{F}$ with $A\in\sigma(\mathcal{D})$. You can show this by showing that the family of sets generated by a countable sub-family of $\mathcal{F}$ is a $\sigma$-algebra containing $\mathcal{F}$.

Community
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Michael Greinecker
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  • Thank you. Such beautiful results! One point, though. You wrote that "If all $\Omega_i$ contains more than one element, their uncountable product is automatically larger than the continuum." It seems like this implies that if $F,G,H$ are cardinals such that $1<F$ and $G,H$ are infinite with $G<H$, then $F^G<F^H$. In particular, if $\aleph_0<H$ then $2^{\aleph_0}<2^H$. Now, i'm a little rusty on my set theory, but isn't this tantamount to the continuum hypothsis? – Evan Aad Apr 13 '13 at 09:46
  • @EvanAad You are right, cardinal exponentiation may fail to be injective. I'll edit my answer. – Michael Greinecker Apr 13 '13 at 09:52