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Can there be two different complex (hence finite) measures whose support is equal? I presume no, as it may defy some separation theorems otherwise. But I need a concrete proof. Any help is hugely appreciated !

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  • $\begingroup$ Multiply one measure by some density which is bounded from away and bounded away from zero. $\endgroup$ – Dirk Dec 10 '18 at 21:19
  • $\begingroup$ What is your definition of support? $\endgroup$ – DisintegratingByParts Dec 10 '18 at 21:44
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The support of a measure $\mu$ is the complement of the largest open set $U$ with $\mu (U)=0$. If $f$ and $g$ are strictly positive integrable functions on $\mathbb R$ and $\mu (A)=\int_A f(x)dx,\nu (A)=\int_A g(x)dx$ then the support of either measure is $\mathbb R$ (because the only open set $U$ of measure $0$ is the empty set). Take $f(x)=e^{-|x|}, g(x)=e^{-2|x|}$, for example to get a counterexample.

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