Is the product measure on the Hilbert brick $I=[0,1]^\mathbb{N}$ outer regular (that is measure of every set is the inf of measures of open sets, containing it)?
2026-05-05 00:07:11.1777939631
Outer Regularity of the Lebesgue measure on the Hilbert brick
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Yes, it is outer regular.
First of all, since $I$ has only countably many factors, the product $\sigma$-field coincides with the Borel $\sigma$-field, that is, $({\cal B}([0,1]))^{\mathbb N}={\cal B}([0,1]^{\mathbb N})$. (see Proposition 8.1.5. on page 256 of [C]). So the product measure is a Borel measure.
Secondly, every finite Borel measure on a Polish space is regular, so you are done. (see Proposition 8.1.10. on page 258 of [C]).
Reference:
[C] Measure Theory (first edition) by Donald L. Cohn, Birkhäuser, 1980.