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I know it is true that every countable set has measure zero, but is the converse true. Is it true that every set of measure zero is countable?

Martin Sleziak
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Roberto
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    You could have gotten the answer from Google in less time it took you to post this. – symplectomorphic May 15 '16 at 05:14
  • Related: http://math.stackexchange.com/questions/459849/examples-of-uncountable-sets-with-zero-lebesgue-measure and http://math.stackexchange.com/questions/1610098/example-of-an-uncountable-dense-set-with-measure-zero – Martin Sleziak May 15 '16 at 06:43
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    If one does not know the answer to this question it might be difficult to predict that in advance @symplectomorphic – Jonathan Hole Mar 14 '20 at 01:11

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No. The Cantor set is probably the easiest example of an uncountable null set.

Of course, there are many others. For instance, every Lebesgue measurable set is a union of a Borel set and a null set.

Martin Argerami
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No, and it's not necessary to jump to the Cantor set (which requires some intense mathematics) in order to see this. Think of a line (such as $\Bbb R \times \{0\}$) in $\Bbb R^2$ - it is a null set, but clearly not countable.

In general, if you have at least a vague intuitive idea of the concept of "dimension", everything that is of dimension $\lneq n$ is a null set in $\Bbb R^n$ (with respect to the usual topological and measurable structures) - and most of these subsets are clearly not countable.

Alex M.
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countable sets are measure zero by definition of measure zero because countable sets we can always use a union of interval with arbitrarily small sum of length to cover it. However, measure zero is not always countable, for example cantor set.

Userkkr
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