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What can be said about $\sigma(T_1 \otimes T_2)$ and $\sigma(T_1) \otimes \sigma(T_2)$, when $T_i$ are topologies that aren't necessary second countable, and $\otimes$ denotes, at the left, the product topology, and at the right, the product $\sigma$-algebra ?

Do you know examples where neither of these is included in the other ?

EDIT : I got an answer of these two questions here : Is "product" of Borel sigma algebras the Borel sigma algebra of the "product" of underlying topologies? Sorry for the duplicate.

I was asking myself if there was a measurable space $(X,\Sigma)$, a topological vector space $(V,\mathbf{T})$ and $\forall i \in \{1,2\}$, $f_i : (X,\Sigma) \rightarrow (V,\mathbf{T})$ measurable, such that $f_1 + f_2$ was $\textbf{not}$ mesurable. This cannot happen if $V$ is second countable, but I don't know many not second countable topological vector spaces...

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To clarify: you want an answer to the question about $f_1 + f_2$, and that's what the bounty is for? –  Nate Eldredge Nov 26 '13 at 16:04
    
Yes, of course ! –  Plop Nov 28 '13 at 13:07

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The sigma algebra $\sigma(\mathbf T_1)\otimes \sigma(\mathbf T_2)$ is always contained in $\sigma(\mathbf T_1\otimes \mathbf T_2)$, essentially by the definition of a product $\sigma$-algebra.

On the other hand, if $(V,\mathbf T)$ is a Hausdorff topological space with cardinality strictly greater than $\mathfrak c$ (the cardinality of the continuum), then the diagonal $\Delta=\{ (x,x);\; x\in V\}$ is a closed set (hence a set in $\sigma(\mathbf\otimes\mathbf T)$) which is not in $\sigma(\mathbf T)\otimes\sigma (\mathbf T)$. A proof of this is outlined for example in Problem 4.1.11 of Dudley's (beautiful) book ``Real analysis and probability".

Now, let $V$ be any Hausdorff topological vector space with cardinality greater than $\mathfrak c$, for example the Hilbert space $\ell^2(I)$, where $I$ is any set with cardinality greater than $\mathfrak c$. Consider the measure space $(X,\Sigma)=(V\times V, \sigma(\mathbf T)\otimes \sigma(\mathbf T))$ and the canonical projections $\pi_1, \pi_2 :V\times V\to V$. These maps are measurable from $(V\times V, \sigma(\mathbf T)\otimes \sigma(\mathbf T))$ into $(V,\sigma(\mathbf T))$, but their difference $f=\pi_1-\pi_2$ is not because $f^{-1}(\{ 0\})$ is equal to the diagonal $\Delta$, which is not in $\sigma(\mathbf T)\otimes \sigma(\mathbf T)$.

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Well done :) ! Thanks a lot. –  Plop Dec 1 '13 at 18:59
    
You're welcome! –  Etienne Dec 1 '13 at 19:12

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