Nov 9, 2012 · $\begingroup$ But proving this is not sigma algebra would require that every infinite set has a countable subset(as much I know proof) but which require Axiom of Countable …

A former advisor of mine said that he thinks a version of Fubini-Tonelli holds in non-$\sigma$-finite spaces (and seemingly had a proof) but he was not entirely convinced by it. Non …

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Sep 13, 2014 · Give an example of a measure space $(\Omega, \mathit{F})$ and a function $\mu$ on $\mathit{F}$ that is additive but not $\sigma$-additive, i.e. $\mu(\cup A_i)= …

Verification that the Borel $\sigma$-algebra on $\mathbb{R}$ is not atomic. 4 Show that the collection of sets for which the inner measure equals the outer measure $\mu_*(A) = …

Recall: the $\bf \Sigma^0_1$ sets are the open sets, the $\bf \Sigma^0_{\alpha+1}$ sets are the sets which can be written as a countable union of complements of $\bf \Sigma^0_\alpha$ sets, …

This won't give you a finite sigma-algebra. The generating set in question is not closed under countable unions, because any set that is a union of infinitely many disjoint left-open-right …

$\begingroup$ The problem becomes much harder if we change it to be: "Find an example that the union of two sigma algebras is an algebra but not a sigma algebra." $\endgroup$ – High GPA

The axiom of choice is not needed to construct a subset of $\Bbb R$ which is not $\sigma$-compact (which I guess is ultimately what this question is about). The irrational numbers form …

I have never heard of the "subtractivity" either. To show that counting measure on $\mathbb{R}$ is not $\sigma$-finite, you just have to note that a set has finite counting measure iff it is finite, …