Talk:Outer measure

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It turns out the second and third requirements together for all sets are incompatible conditions

I'm not sure this is true. Counting measure is a counterexample. I think if you include the first requirement as well (which should imply translation invariance) then the measure cannot be defined for all subsets. Also, is outer measure something which can be defined on (the power set of) any set?, or just Euclidean spaces? Revolver 17:17, 8 Mar 2004 (UTC)
Counting measure is a counterexample + it is translation invariant, but, of course, it does not measure length, area, volume, et. c. -- Leocat 18:36, 31 October 2006 (UTC)

[edit] Measurable sets

In the section "Formal definitions" you state: "A subset E of X is φ-measurable iff for every subset A of X : \varphi(A)= \varphi(A\cap E) + \varphi(A\setminus E). "

What if the set A is not measurable?

-- Leocat 18:36, 31 October 2006 (UTC)

The definition (due to Caratheodory) explicitly quantifies over all subsets of X. What's the problem?--CSTAR 18:45, 31 October 2006 (UTC)

If the set A is not measurable and contained in E, then \varphi(A\cap E) is not defined. -- Leocat 11:07, 1 November 2006 (UTC)

The definition says
An outer measure is a function defined on all subsets of a set X
\varphi: 2^X \rightarrow [0, \infty]
such that some additional conditions (countable subadditivity etc.) hold. Therefore \varphi(A\cap E) is defined for any subset A of X. This is the standard definition; please look at any of the cited refernces. --CSTAR 15:06, 1 November 2006 (UTC)