Suppose $\nu$ is a regular signed or complex Borel measure on $\mathbb R^n$, m 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 f is an extended m-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.

PS: A Borel measure $\nu$ on $\mathbb R^n$ is regular if $\nu(K)<\infty$ for every compact K. 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.


  • $\begingroup$ Is this something you have seen in a textbook and are trying to work out? or a generalization of something you have seen? $\endgroup$
    – Yemon Choi
    Mar 20, 2011 at 9:45
  • $\begingroup$ As phrased, this does not really look appropriate for MO; but perhaps I am failing to notice some subtlety here... $\endgroup$
    – Yemon Choi
    Mar 20, 2011 at 9:46
  • $\begingroup$ I don't understand your comment. $\endgroup$
    – zzzhhh
    Mar 20, 2011 at 9:49
  • $\begingroup$ Does my question hurt you? $\endgroup$
    – zzzhhh
    Mar 20, 2011 at 9:50
  • 2
    $\begingroup$ He means it looks like a homework-type problem (which is inappropriate here) and not a research-type problem (which is appropriate here). $\endgroup$ Mar 20, 2011 at 13:18

1 Answer 1


If $\nu\bot\lambda$, then $|\nu+\lambda|=|\nu|+|\lambda|$


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