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 October 14th, 2013, 01:10 AM #1 Senior Member   Joined: Sep 2010 From: Germany Posts: 153 Thanks: 0 show that $\mu$ is a measure let $\bar{{\mathcal{M}}}=\{E\cup F|E\in\mathcal{M}|F\subset N|N\in\mathcal{N}\}$ be a sigma algebra where $\mathcal{N}=\{N\in\mathcal{M}|\mu(N)=0\}$ how to show that $\bar{\mu}(E\cup F)=\mu(E)$ is a measure, where $\mu$ is a measure on $\mathcal{M}$ $\bar{\mu}(\emptyset\cup \emptyset)=\mu(\emptyset)=0$? countable unions?
 October 14th, 2013, 02:21 AM #2 Senior Member   Joined: Feb 2013 Posts: 281 Thanks: 0 Re: show that $\mu$ is a measure mu is a measure on M, so mu(empty) = 0 I give a proof for a union of two disjoint sets. I hope you can extend the proof for countable pairwisely disjoint sets. I will detone the new measure by nu. We have to show that nu(EuF u E'uF') = nu(EuF) + nu(E'uF'), where EuF and E'uF' are disjoint sets. The left side is equal with mu(E) + mu(E') by definiton of nu. The right hand side nu(EuF u E'uF') = nu(EuE' u FuF') = nu(EuE' u F'') = mu(EuE') = mu(E) + mu(E') where F'' = FuF' < NuN' = N''. So the two sides are equal.
 October 14th, 2013, 02:30 AM #3 Senior Member   Joined: Sep 2010 From: Germany Posts: 153 Thanks: 0 Re: show that $\mu$ is a measure ah I see it now. Thank you!

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### show that u(E) u(F) =u(E U F) u(E n F) where u is measure of sigma algebra

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