Let $n$ be a positive integer. It is well-known that $\mathbb{R}^n$ cannot be non-trivially partitioned into open sets, since it is connected.

Let $\frak P$ be a partition of $\mathbb{R}^n$ into closed sets and assume $\mathbb{R}^n\notin{\frak P}$ (that is, ${|\frak P|}>1$). Let ${\frak P}_0\subseteq {\frak P}$ consist of the elements of ${\frak P}$ that have Lebesgue measure $0$. Is it necessarily true that $|{\frak P}_0| = 2^{\aleph_0}$?

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    $\begingroup$ Just a small remark: actually all elements of $\mathfrak{P}$ are closed, thus Borel measurable, hence Lebesgue measurable, right? $\endgroup$ – Jochen Glueck Jun 8 '19 at 6:29
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    $\begingroup$ @PietroMajer how do we make such a function? $\endgroup$ – Fedor Petrov Jun 8 '19 at 7:49
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    $\begingroup$ Not necessarily true. It is consistent that $2^{\aleph_0}$ is arbitrarily large and every Polish space can be partitioned into $\aleph_1$ non empty closed sets. See Theorems 3 and 4 in A. Miller, Covering $2^{\omega}$ with $\omega_1$ disjoint closed sets, here math.wisc.edu/~miller/res/cov.pdf $\endgroup$ – Ashutosh Jun 8 '19 at 9:25
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    $\begingroup$ I think several of these comments should be posted as answers. $\endgroup$ – Joel David Hamkins Jun 8 '19 at 9:52
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    $\begingroup$ I see, the quotient is not necessarily $T_2$! $\endgroup$ – Pietro Majer Jun 8 '19 at 10:02

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