Prove that $\mathbb{R}=\sqcup_{i\in \mathfrak{c}}A_i, \mu^*(A_i)>0$

Question

Prove that $\mathbb{R}$ can be partitioned into continuum disjoint sets with positive outer measure.

Are you sure this is true? – Rushabh Mehta Nov 08 '19 at 23:20 — Rushabh Mehta, Nov 08 '19 at 23:20
Perhaps $$ \mathbb R = \sqcup_{n\in\mathbb Z}[n-1,n) $$ – Math1000 Nov 08 '19 at 23:43 — Math1000, Nov 08 '19 at 23:43

bof · Accepted Answer · 2019-11-09T02:11:42.907

1

Partition $\mathbb R$ into continuum many disjoint sets (so-called Bernstein sets), each of which has nonempty intersection with every uncountable closed set. (A set which meets every uncountable closed set while containing no uncountable closed set is called a Bernstein set.) This is easily done by transfinite induction, since there are just continuum many uncountable closed sets, and every uncountable closed set contains continuum many points.

edited Nov 09 '19 at 02:11

answered Nov 09 '19 at 01:26

bof

78,265

neat +1. do they all have infinite outer measure? – Tim kinsella Nov 09 '19 at 01:38
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They all have full outer measure. The complement of each has inner measure zero, as it contains no uncountable closed set. – bof Nov 09 '19 at 01:42
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related https://en.wikipedia.org/wiki/Bernstein_set and https://math.stackexchange.com/q/169714 – Mirko Nov 09 '19 at 02:03

Prove that $\mathbb{R}=\sqcup_{i\in \mathfrak{c}}A_i, \mu^*(A_i)>0$

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