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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