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$\begingroup$ Even though this isn't strictly an answer to the question, I'm going to accept it on the reasoning that it'll probably be most useful to someone visiting this thread in the future. I don't like the thought of somebody reading just the first couple of answers and missing this gem. $\endgroup$
– Anton Geraschenko
Commented Oct 29, 2010 at 2:32
1

$\begingroup$ In fact Shelah proved that “ZF+CC+all sets of real numbers are measurable” is equiconsistent to the existence of an inaccessible cardinal. Without countable choice you could simply choose a model of ZF where $\mathbb{R}$ is a countable union of countable sets. $\endgroup$
– The User
Commented May 20, 2013 at 20:24
$\begingroup$ @The User: Yes, you are absolutely right. Thanks for pointing this out. $\endgroup$
– Stefan Geschke
Commented Oct 19, 2015 at 16:47

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