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Why is the Borel $\sigma$-Algebra on a subset of $\mathbb{R}^d$ said to be incomplete wrt. to the Lebesgue measure?

As I understand a complete measure space contains all subsets of all sets of measure zero from the $\sigma$-Algebra. Any subset of a set with measure zero in the Borel $\sigma$-Algebra of a subset of $\mathbb{R}^d$ is of dimension $d-1$ which means it is closed, so again measurable.

Perelman
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    I think it's overlapped with say https://math.stackexchange.com/questions/575396? – Liding Yao Mar 21 '24 at 16:23
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    Hi, welcome to MSE! I think the mistake in your reasoning is "any subset of a set with measure zero must be closed". This is not true - consider for example the measure-zero set $\Bbb R \times {0} \subseteq \Bbb R^2$. This has many non-closed subsets - for example, $(0, 1) \times {0}$, as well as non-Borel subsets as explained at the link in the above comment. Possibly you're conflating "closed" and "not open". – Izaak van Dongen Mar 21 '24 at 16:29
  • @IzaakvanDongen I see thank you. But if $d=1$ then any zero set consists of countably many points right? so any subset is a set of measure zero again, right? – Perelman Mar 21 '24 at 16:42
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    It's not true that any null set in $\Bbb R$ consists of countably many points. The standard example is the Cantor set! – Izaak van Dongen Mar 21 '24 at 16:44
  • Can you please define this "dimension" you speak of? – Sassatelli Giulio Mar 21 '24 at 16:50
  • @SassatelliGiulio i meant that in a informal way. What I was saying was if a set has zero volume or zero area then the object must be a hyperplane (so one dimension less). For instance something that has finite volume must be flat ect. – Perelman Mar 21 '24 at 19:49
  • @IzaakvanDongen So when formulating the standard convergence result of lebesgues dominated convergence theorem in for example $(\mathbb{R},\mathcal{B}, \lambda)$ we also need to impose always that the limit function is measurable? Since we can not guarantee its measurability if the underlying measure space is incomplete. – Perelman Mar 21 '24 at 19:51
  • I don't think so. A pointwise limit of Borel measurable functions is automatically Borel. For a measure space, "incomplete" doesn't have much to do with "measurability of functions is preserved under limits". – Izaak van Dongen Mar 22 '24 at 14:41

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