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Topic: outer measure (Read 2955 times) |
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MonicaMath
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outer measure
« on: Sep 27th, 2009, 9:38pm » |
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is there an example for a disjoint sequense of sets {A_k} that satisfies the countably subadditive (strictly < ) property of outer measure ??
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Eigenray
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Re: outer measure
« Reply #1 on: Sep 28th, 2009, 12:26am » |
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A simple example is to take any set with more than one element, and define the outer measure of any non-empty set to be 1.
If you mean an example for the Lebesgue outer measure, consider a Vitali set V. All translations of V have the same outer measure c, so countable subadditivity implies c > 0. But the sum is then infinite, giving strict inequality, no matter the actual value of c (which depends on the choice of V).
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MonicaMath
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Re: outer measure
« Reply #2 on: Sep 28th, 2009, 11:31pm » |
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OK, thank you.
But I'm wondering .... does the Vitali set has measure 1 ?? and if so does this contradicts that it is a subset of [0,1] which has measure 1 also?
thank you
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Eigenray
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Re: outer measure
« Reply #3 on: Sep 29th, 2009, 4:01am » |
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on Sep 28th, 2009, 11:31pm, MonicaMath wrote:OK, thank you.
But I'm wondering .... does the Vitali set has measure 1 |
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The Vitali set V is not measurable. The union of countably many disjoint translates of V has positive, finite measure, so there is no way for countable additivity to hold. But V is outer measurable. It has positive outer measure, which demonstrates strict subadditivity.
The actual outer measure of V depends on V, but it can be any number in (0,1]. See this post.
Quote:| and if so does this contradicts that it is a subset of [0,1] which has measure 1 also? |
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I don't know what you mean. There are many subsets of [0,1] which have measure 1, for example [0,1] with countably many points removed. More generally one can remove any set of measure 0, like the Cantor set.
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