Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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...

share|cite|improve this question
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
up vote 2 down vote accepted

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 $\sigma(\mathbf T)\otimes \sigma(\mathbf T)$.

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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.