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On $(\mathbb{R}, \mathcal{B})$ given any finite measure $\mu$ the sets of the form (continuity sets) $$\{A \in \mathcal{B} : \mu(\partial A) = 0\}$$ generate the Borel $\sigma$-algebra $\mathcal{B}$. The same is true in $\mathbb{R}^n$ with some measure $\mu$ in $(\mathbb{R}^n, \mathcal{B}_{\mathbb{R}^n})$. "Decomposition of Multivariate Probability", Roger Cuppens or https://math.stackexchange.com/questions/1634436/sigma-field-generated-by-the-continuity-sets-of-a-measure/1634839.

I'm wonder if this is just an easy result:

Consider $(\mathbb{R}^\infty, \mathcal{B})$, where $\mathcal{B}$ is the $\sigma$-algebra generated by the product topology, which agree with the one generated by cylindrical sets: Given some finite measure $\mu$ on $(\mathbb{R}^\infty, \mathcal{B})$ is enough to use cylindrical based on continuity sets? i.e: by sets of the form $$\{(x_k): \pi_n((x_k)) \in A, n >0, \mu_n(\partial A)=0, A \in \mathcal{B}_n\}$$ where $\pi_n$ projects $(x_1, x_2, x_3, \cdots )$ to $(x_1, x_2, \cdots, x_n)$ and $\mu_n$ is the push of $\mu$ through $\pi_n$?.

Similar idea would work for $X^{[0,1]}$, when $X$ is a compact metric space with a finite measure defined on the cylindrical $\sigma$-algebra?

I feel the argument given in the answer linked above is enough, since this set will contain a base for the topology of the projections. I haven't seen this anywhere and it seems quite useful, is there any reason? any reference actually using this?

Bye.

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  • $\begingroup$ There is some typo above, in your description of cylinder sets, $\mu_n(\partial A)$ is not a condition (yet). Also, I'm confused about what you mean by $X^{[0,1]}$, which is not even metrizable under the obvious interpretation (all functions from $[0,1]$ to $X$, endowed with the product topology) if $X$ is nondegenerate. $\endgroup$ – Michael Greinecker Oct 31 '16 at 9:43
  • $\begingroup$ @MichaelGreinecker $\mu_n$ is defined from $\mu$ using the projections to $\mathbb{R}^n$, so given $\mu$ on the cylindrical $\sigma$-algebra, such condition make sense. It is possible to see $\mu_n$ as a measure on $\mathbb{R}^n$ or as a measure on $\mathbb{R}^\infty$ restricted to $\sigma(\pi_n)$. In the second case, the usual cylindrical $\sigma$-algebra is only a sub $\sigma$-algebra of the one generated by the product topology, I'm interested in checking if the former is constructable using the "continuity sets" as base of the cyliders. $\endgroup$ – Jorge E. Cardona Oct 31 '16 at 12:41
  • $\begingroup$ There is still a plain typo the way you write your cylinder sets; I presume you want to have $\mu_n(\partial A)=0$. Since continuity-sets are defined in terms of a topology, you would still have to specify which topology to put on $X^{[0,1]}$. $\endgroup$ – Michael Greinecker Oct 31 '16 at 14:16
  • $\begingroup$ @MichaelGreinecker Oh, yes, thank you. $\endgroup$ – Jorge E. Cardona Oct 31 '16 at 14:18
  • $\begingroup$ @MichaelGreinecker Is it not enough to have a topology (hence the Borel $\sigma$-algebra) on $X$ to be able to define the cylindrical $\sigma$-algrebra? Having the product topology on $X^{[0,1]}$ would allow us to construct a bigger $\sigma$-algebra no?, I'm so far only interested in the former case. $\endgroup$ – Jorge E. Cardona Oct 31 '16 at 14:49
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Fix a finite measure $\mu$ on $\mathbb{R}^\infty$. For each $n\in\mathbb{N}$, there are at most countably many $x\in\mathbb{R}$ such that $$\mu\big(\mathbb{R}^{n-1}\times\{x\}\times\mathbb{R}\times\mathbb{R}\times\ldots\big)>0.$$ So for each $n$ there is a countable dense set $D_n$ such that $$\mu\big(\mathbb{R}^{n-1}\times\{x\}\times\mathbb{R}\times\mathbb{R}\times\ldots\big)=0$$ for all $x\in D_n$. Now the Borel-$\sigma$-algebra on $\mathbb{R}^\infty$ is generated by cylinder sets of the form $$\bigcap_{n\in F\\ a_n\in D_n\\ b_n\in D_n}\pi_n^{-1}\big([a_n,b_n]\big)$$ for some finite set $F\subseteq\mathbb{N}$. These cylinder set are all $\mu$-continuity sets, are closed under finite intersections, and there are only countably many of them.


It is also possible to metrize $\mathbb{R}^\infty$ by a metric $\rho$ given by $$\rho\big(\langle x_n\rangle,\langle y_n\rangle\big)=\sum_{k=1}^\infty 1/2^k\frac{d(x_k,y_k)}{1+d(x_k,y_k)}$$ with $d(a,b)=|b-a|$. Fix a countable dense set $D\subseteq\mathbb{R}$. For each $x$, there are only countably many $r$ such that the open ball $B_r(x)$ is not a continuity set. Let $G_x$ be a countable dense set of radii. Then the family of finite intersections of Balls $B_r(x)$ with $x\in D$ and $r\in G_x$ forms a countable family of $\mu$-continuity sets that generates the Borel $\sigma$-algebra and is closed under finite intersections. Instead of $\mathbb{R}$, one can use any separable metric space.

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