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Topic: outer measure countably additive
Replies: 19   Last Post: Jul 3, 2013 12:20 PM

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RGVickson@shaw.ca

Posts: 1,645
Registered: 12/1/07
Re: outer measure countably additive
Posted: Sep 23, 2010 9:40 PM
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On Sep 23, 7:32 am, Anonymous wrote:
> I have a problem to prove that m* is countably additive on the class
> of open subsets of [0,1].  Any hints on how to get started?


I assume that m* means Lebesgue outer measure. Are you allowed to use
the following three properties? (1) Sub-additivity m*(union_{i=1}
^{inf} A_i) <= sum_{i=1}^{inf} m*(A_i); (2) monotonicity A \subset B --
> m*(A) <= m*(B); and (3) finite additivity m*(A union B) = m*(A) +
m*(B) for disjoint A and B. If so, let {A_i} be disjoint intervals,
and let's look at the case with union A_i \subset J for some finite
interval J (just to make sure we are dealing with a finite value of
m*). We have m*(union A_i) <= sum m*(A_i), so we need only establish
the opposite inequality. For finite n we have union_{i=1}^{inf} A_i
contains union_{i=1}^{n} A_i, so m*(union_{i=1}^{inf} A_i) >=
m*(union_{i=1}^{n} A_i) = sum_{i=1}^{n} m*(A_i), by finite additivity.
Now let n --> inf. If you are not allowed to use properties (1)--(3),
you can start by trying to first establish those properties. A Google
search under 'outer measure' turns up lots of relevant web pages.

R.G. Vickson



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