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Wikipedia states that

the Lebesgue measure \lambda is an extension of "the" Borel measure which possesses the crucial property that it is a complete measure (unlike the Borel measure).

However I have read that for every Lebesgue-measurable set a subset can be found, which is not measurable (some kind of Vitali set inside a measurable set, if I'm not mistaken). So I could take a null set, e.g. the Cantor set, and with help of the axiom of choice I can create some sort of Vitali set, which is a subset of the Cantor set. This, however would contradict Wikipedia's claim that the Lebesgue measure is a complete measure.

Pretty new to the whole measure theory stuff, so I guess my mistake is a pretty obvious one. Any help is greatly appreciated :)

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You have the wrong definition of complete. A measure is complete if whenever two sets are almost equal (i.e. differ by a set of measure zero), and one of them is measurable, then the other is measurable. Another way to say this is that your $\sigma$-algebra has all the available sets of measure zero.

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  • $\begingroup$ I knew I shoudn't learn from Wikipedia, but this seems to be completely different to how Wikipedia defines complete: en.wikipedia.org/wiki/Complete_measure : "In mathematics, a complete measure [...] is a measure space in which every subset of every null set is measurable (having measure zero).", or am I missing something? $\endgroup$ – Peter Oct 27 '13 at 11:03
  • $\begingroup$ No, this the same, and I now see your confusion. Where did you read that a Vitali set can be constructed from every measurable set? We need that reference to set where you are going wrong, because I'm fairly certain this assertion is incorrect. $\endgroup$ – user452 Oct 27 '13 at 11:06
  • $\begingroup$ analysis.math.uni-kiel.de/schuett/masstheo.pdf (danger: German) on site 58 (point III), but dumb me just realized that this only holds true for sets whose measure is > 0... $\endgroup$ – Peter Oct 27 '13 at 11:11
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    $\begingroup$ Right, and the reason is that if you are constructing your Vitali set from an interval of measure $1$, say, then the Vitali set can have outer measure $1$, but so can its complement in your interval, and so you never obtain additivity. $\endgroup$ – user452 Oct 27 '13 at 11:15
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You write

However I have read that for every Lebesgue-measurable set a subset can be found, which is not measurable (some kind of Vitali set inside a measurable set, if I'm not mistaken).

What you should have read is that every set of positive outer Lebesgue-measure has an immeasurable subset.

Clearly every subset of a null set has outer measure 0.

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