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June 18

Measure Theory and non-measurable (caratheodory-measurability) set

If I have a set X = {1,2}, and a measure m on this set given by

m({}) = 0
m({1}) = m({2}) = m({1,2}) = 1

then by the Caratheodory definition of measurability, the set {1} is not measurable because

1 = m({1,2}) ≠ m({1,2} ∩ {1}) + m({1,2} - {1}) = 2

.

But m({1}) = 1 is well defined. Is this like that for all non-measurable sets by the caratheodory definition of measurability? Or is there a criteria when the actual value of the measure can be determined regardless of the non-measurability of the set?

In the book I use for study for example, the vitali-set V is given as an example of a non-measurable set. But there there is a proof by contradiction, which says that m(V) (there using the lebesgue measure) does not make any sense, and assigning any value to it results in a contradiction.

--helohe (talk) 18:58, 18 June 2013 (UTC)[reply]

The Vitali set has a well-defined outer measure with respect to the Lebesgue measure, but it is not measurable. I'm dubious of your example above. The gadget m there is not an additive measure, but the usual Caratheodory construction starts with an additive measure on a ring of sets. Sławomir Biały (talk) 21:27, 18 June 2013 (UTC)[reply]

In the definition we use the only additivity requirement is: m(A) ≤ ∑ m(A_k) where A ⊆ ∪A_k

This requirement should be satisfied as 0 ≤ 1 ≤ 1

--helohe (talk) 21:35, 18 June 2013 (UTC)[reply]

That's usually called a "submeasure" rather than a measure (though in contexts where you're mainly interested in submeasures, sometimes they're referred to as "measures" for short, at least in informal communications). A measure, strictly speaking, has exact additivity for countable sequences of pairwise disjoint sets. --Trovatore (talk) 22:55, 18 June 2013 (UTC)[reply]

In what definition do you only use this? The definition of measure (mathematics) requires additivity, not just subadditivity. Outer measures are generally only subadditive. Your m is actually the outer measure associated to the (true) measure

\mu (\emptyset )=0,\mu (\{1,2\})=1

. Sławomir Biały (talk) 11:51, 19 June 2013 (UTC)[reply]

I use the definition from the lecture notes (in german, but should be easy to understand): http://www.math.ethz.ch/~struwe/Skripten/AnalysisIII-SS2007-18-4-08.pdf --helohe (talk) 22:35, 19 June 2013 (UTC)[reply]

Well, it is certainly possible that some workers somewhere require only subadditivity in the definition of a measure; that's not wrong a priori; it's just not the most usual definition. But the fundamental point is that the definition of measures relevant to the Caratheodory criterion requires full additivity. --Trovatore (talk) 22:49, 19 June 2013 (UTC)[reply]

The Caratheodory criterion applies to a (subadditive) outer measure, such as m above, and produces its restriction which is an (additive) measure. It makes no sense to require m to be additive, since then the criterion would be satisfied for all sets.—Emil J. 11:28, 20 June 2013 (UTC)[reply]

I think the confusion here is that the source referenced calls a measure what other sources would call an outer measure. This was certainly my own reason for objecting to the example given. Sławomir Biały (talk) 11:46, 20 June 2013 (UTC)[reply]