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I have a problem with Proposition 7.2.10 in Bogachev's Measure Theory Volume II book on page 77 (I have link to my drive with that book https://drive.google.com/file/d/1CgzgWhEiNPU1vy0YkyiVjGLH3Qiq1aCL/view). Supposedly, from that proof one can deduced that $\text{supp}\,\mu$ is separable if $\mu$ is $\tau$-additive Borel measure on metric space $(X, d)$ and $\mu$ has full support ($\mu(X \setminus \text{supp}\,\mu) = 0$). I don't see it, and also I don't know what is $\Gamma$ in this proof. I think, it is not specified. Maybe anyone knows answers for my questions or at least one of them? I'd appreciate.

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  • $\begingroup$ "Therefore, we obtain a family $\Gamma$ of open sets of $\mu$-measure zero such that their union has a positive $\mu$-measure." $\endgroup$ Commented Mar 21, 2020 at 20:37
  • $\begingroup$ Yeah, but I can's see why $\Gamma$ exist. From what we know it ? $\endgroup$
    – elsnar
    Commented Mar 21, 2020 at 20:39

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Following Bogachev's notation and terminology, let $S_\mu$ be the intersection of all closed sets of full $\mu$-measure, and say $\mu$ "has support" if $S_\mu$ itself has full measure. A useful fact is that $x \in S_\mu$ if and only if every open neighborhood of $x$ has positive measure.

Now suppose $\mu$ is a finite measure, as in the proof. If $S_\mu$ were not separable, then as Bogachev notes, it would be possible to find an uncountable collection of disjoint open balls centered at points of $S_\mu$. By the useful fact above, they must all have positive measure, and in particular, for some $n$, there are infinitely many with measure at least $1/n$, contradicting the finiteness of $\mu$.

As for $\Gamma$, at this point we are supposing that $\mu$ is a measure without support, so that $\mu(S_\mu^c) > 0$. Now by the fact above, every $x \in S_\mu^c$ has an open neighborhood $U_x$ of measure zero; necessarily $U_x \subset S_\mu^c$ and without loss of generality $U_x$ is a ball. Take $\Gamma = \{ U_x : x \in S_\mu^c\}$. Their union is $S_\mu^c$ which has positive measure.

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  • $\begingroup$ Thanks! You made it understable. Author's reasoning was too fast for me. $\endgroup$
    – elsnar
    Commented Mar 21, 2020 at 21:38

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