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Let $\mu $ be a measure on $(X,\mathcal{P}(X))$ where $|X|=\aleph_1$. Show that there exists a function $f:X\longrightarrow [0,\infty)$ and a $0-\infty$ measure $\nu$ on $(X,\mathcal{P}(X))$ such that: $$ \mu\left(E\right)=\sum_{x\in E}f\left(x\right)+\nu\left(E\right) $$ For all $E\subseteq X$.

Note: a $0-\infty$ measure is a measure that only takes the value $0$ and $\infty$.

Been thinking about it for a while and still not sure where to start. Any help would be highly appreciated. Thanks in advance.

EDIT:

I proved that $$ \mu=\mu_{0}+\nu $$ Where $$ \mu_{0}\left(E\right)=\sup\left\{ \mu\left(F\right)\thinspace:\thinspace F\subseteq E,\thinspace F\in\mathcal{M}\thinspace\mu\left(F\right)<\infty \right\} $$

and $\nu$ is a $0-\infty$ measure.

Need to find a way to show that in the particular case of $|X|=\aleph_1$ this $\mu_0$ can be shown a sum of the function $f(x)$over the set $E$.

DirichletIsaPartyPooper
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    Is $\mu$ sigma-finite here or something? Because if we take $\mu = \infty \delta_0$, then $f$ is not uniquely defined, as it can take any value at $0$. – Damian Pavlyshyn Dec 12 '22 at 02:31
  • Assuming some set theoretic axioms (consistent with ZFC) https://math.stackexchange.com/a/209533/405572, we can take $\mu$ to be Lebesgue measure on $([0,1], \mathcal P([0,1]))$. Then $\nu$ is not going to take the value $\infty$, so it must be identically $0$, so $\mu(E) = \sum_{x\in E} f(x)$. What can $f$ possibly be? I don't think the claim is true. – D.R. Dec 12 '22 at 02:45
  • @D.R. The Lebesgue measure is not defined on $\mathcal{P}([0,1]).$ – Ramiro Dec 12 '22 at 04:12
  • @Ramiro that's why I explicitly said/linked "Assuming some set theoretic axioms (consistent with ZFC) https://math.stackexchange.com/a/209533/405572". If this problem has a proof in ZFC, it should also hold in ZFC + CH + large cardinal axioms (one of which is the existence of an extension of Lebesgue measure to $\mathcal P([0,1])$, as Asaf said in the linked answer) – D.R. Dec 12 '22 at 05:29
  • @D.R. Ok. I see your point. Thanks. – Ramiro Dec 12 '22 at 05:39
  • @DamianPavlyshyn You are right. I fixed the post without the requirement of uniqueness – DirichletIsaPartyPooper Dec 12 '22 at 21:58
  • @Ramiro Any chance you have an idea to prove existance? – DirichletIsaPartyPooper Dec 13 '22 at 11:41
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    As @D.R. pointed out, assuming the consistency of ZFC, this result can not be proved. Why? Because, if ZFC is consistent, we can add some axioms that are known to be consistent with ZFC and with this extended set of axioms, we can extend the Lebesgue measure to $\mathcal{P}([0,1])$. Let us take such extension as $\mu$. Then $\mu([0,1])=1$, so $\nu \equiv 0$. Also, for all $x\in [0,1]$, $\mu({x})=0$ , so $f\equiv0$. But $\mu$ is not identically zero. – Ramiro Dec 14 '22 at 01:17

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