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I need to show that there are distinct $\sigma$-finite measures in $B(\mathbb{R})$ that are the same in the open sets. Im not really seeing a good example cause every open set is a countable union of compact sets, so the sum has to be the same , but they have to interchange themselves. Any help is appreciated.

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For a real number $x\in\mathbb R$ take $\mu_x(A)=\int_{A}\frac{\mathrm dt}{t^2}+x\delta_0(A)$

Then, $\mathbb R=\{0\}\cup \bigcup_n\left(]-\infty,-\frac1n[\cup]\frac1n,+\infty[\right)$ and each subset has finite measure so $\mu_x$ is $\sigma$-finite. Also it is clear that every $\mu_x$ has the same value on an open set which does not contain $0$, and is infinite as soon as $0$ is in the open set, so that all $\mu_x$ coincide on all open subsets but are not equal.

Thank you Kavi Rama Murthy for noticing me that my former answer was wrong.

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I think i found an example. If we consider $\mu(A) = \# (\mathbb{Q} \cap A)$ and $\nu(A) = \# (A \cap(\mathbb{Q}\cup \sqrt{2}))$.

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  • $\begingroup$ Are these measures $\sigma$-finite? $\endgroup$ – EpsilonDelta May 13 at 12:41
  • $\begingroup$ Yeah if we consider R to the union of the $\q_{ns}$ and the irrationals. $\endgroup$ – Pedro Santos May 13 at 12:42
  • $\begingroup$ No they have 0 and 1 i think $\endgroup$ – Pedro Santos May 13 at 12:44
  • $\begingroup$ Surely we have $\mu(\mathbb{Q}) = |\mathbb{Q}| = \infty$? $\endgroup$ – EpsilonDelta May 13 at 12:45
  • $\begingroup$ Yes we do , but we consider X to be the countable union of $q_{n}$ and each one of them has measure 1. $\endgroup$ – Pedro Santos May 13 at 18:37

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