
Re: outer measure countably additive
Posted:
Sep 23, 2010 9:40 PM


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) Subadditivity 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

