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The Lesbegue measure on the real line is defined on the class of Lesbegue-measurable sets, whose elements satisfy the Caratheodory condition.

Here what I am curious about is that if it is possible to 'extend' the domain of Lesbegue measure to some properly bigger subset of the real line's power set and define a measure on the set which is just the Lesbegue measure when restricted to the domain of Lesbegue measure and satisfies the countable additivty.

If, not possible, could anyone explain the reason?

Keith
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1 Answers1

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You'd have to lose something that makes the Lebesgue measure what it is.

The Lebesgue measure has a few distinguishing properties: 1) Countable additivity 2) Preservation of measure under congruences (translations, rotations, reflections) 3) Volume of the unit cube is 1 4) Complete-ness

The Lebesgue measure is unique by uniqueness of extension of Borel measure to a complete measure.

Batman
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  • I read the folland real analysis and yes, Lesbegue measure is the completion of Borel measure. But, Folland does not say about the uniqueness of enlarging the domain. So, what I am curious about seems to be reduced to this : Is there some other completion of Borel sigma algebra on the real line rather than the class of Lesbegue measurable sets. – Keith Mar 24 '15 at 15:54
  • Well, completion has a very specific meaning, so I don't think you mean completion. The Lebesgue measure is the unique measure on $\mathbb{R}^d$ satisfying the properties I stated. Maybe this question is what you're looking for? – Batman Mar 24 '15 at 16:06