most recent 30 from http://mathoverflow.net 2013-05-19T01:32:32Z http://mathoverflow.net/feeds/question/58960 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/58960/a-question-about-regular-signed-or-complex-borel-measure-under-lrn-decomposition zzzhhh 2011-03-20T09:13:07Z 2011-03-20T10:49:19Z

<p>Suppose $\nu$ is a regular signed or complex Borel measure on $\mathbb R^n$, <em>m</em> is the Lebesgue measure on the class of Borel sets $\mathcal B_{\mathbb R^n}$ and the Lebesgue-Radon-Nikodym decomposition of $\nu$ is $d\nu=d\lambda+fdm$ where <em>f</em> is an extended <em>m</em>-integrable function when $\nu$ is a signed measure or $\in L^1(m)$ when $\nu$ is a complex measure and $\lambda\bot m$. Prove that $d|\nu|=d|\lambda|+|f|dm$ where the notation $|\bullet|$ represents total variation. </p> <p>PS: A Borel measure $\nu$ on $\mathbb R^n$ is regular if $\nu(K)&lt;\infty$ for every compact <em>K</em>. From this definition we have $\nu(E)=\inf \{\nu(U)|U{\;\rm{ open},}U\supseteq E\}$. A signed or complex Borel measure on $\mathbb R^n$ is regular if $|\nu|$ is regular.</p> <p>Thanks!</p>

http://mathoverflow.net/questions/58960/a-question-about-regular-signed-or-complex-borel-measure-under-lrn-decomposition/58962#58962 Syang Chen 2011-03-20T10:49:19Z 2011-03-20T10:49:19Z

<p>If $\nu\bot\lambda$, then $|\nu+\lambda|=|\nu|+|\lambda|$</p>