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Given $\nu_1,\nu_2$ finite signed measures, is there a way to prove $|\nu_1 + \nu_2| \le |\nu_1| + |\nu_2|$ without resorting to the fact that for a general signed measure $\nu$, $$|\nu|(E) = \sup \sum_{i=1}^{n}|\nu(E_i \cap E)|,$$ where the $E_i$ are disjoint, measurable?

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  • $\begingroup$ you want to prove the subadditivity of the total variation without using its definition? $\endgroup$
    – Lolman
    Nov 15, 2014 at 16:23
  • $\begingroup$ I want to prove it directly from the Jordan decomposition definition in wisher's post. I can't find a way to do it directly from this. $\endgroup$
    – user186541
    Nov 15, 2014 at 16:52

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One way you may want to try is using $\nu = \nu^+-\nu^-$ and $\nu=\nu^++\nu^-$, the Jordan decomposition. This could simplify the problem a little bit. But as @Lolman suggest, you can not actually avoid the definition of total variation, as the above identity is proved by the definition of total variation.

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