We know for the Borel $\sigma$-algebra that each singleton set is measurable. I was working on the problem of proving that each infinite $\sigma$-algebra has uncountably many members. My solution went something like this: if $\sigma$ were countable, intersect all of the measurable sets which contain $x$. I derive a contradiction from this. Of course, $\sigma$-algebras aren't necessarily closed under arbitrary intersection, so it might not make sense to talk about "smallest measurable set" containing a point. My question is what conditions can we put on the measure space so that it does make sense to talk of the smallest measurable set containing a point?
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Przemysław Scherwentke
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asked Jul 18, 2013 at 20:21
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$\begingroup$ I assume by arbitrary intersection you mean uncountable? Because $\sigma$-algebras are certainly closed under countable intersections by definition. $\endgroup$– KeatonJul 18, 2013 at 20:40
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$\begingroup$ If the whole sigma-algebra were countable, then the sub-collection of measurable sets that contain $x$ would also be countable. So its intersection would be measurable. $\endgroup$– GEdgarJul 18, 2013 at 21:02
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$\begingroup$ Right. That's not the contradiction I get. I conclude that any measurable set that contains $x$ must contain the smallest set $E_x$. But then this decomposes the measure space into "atoms", which each nonempty measureable set is just the union of atoms $E_{x_i}$. If there were finitely many atoms, our $\sigma$-algebra is finite. If it is countable, then the measurable sets can be put into a bijective correspondence with the set of sequences of $0$s and $1$s. Hence, it would be uncountable. $\endgroup$– Alex LapanowskiJul 18, 2013 at 21:05
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$\begingroup$ The fact that every sigma-algebra is either finite or uncountable is often proved introducing such atoms--but this is not your question, right? $\endgroup$– DidJul 20, 2013 at 8:38
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$\begingroup$ It's not my question. But the idea of picking a smallest measurable set which contained a specific point came up naturally in its solution. Its what led me to think of the question. $\endgroup$– Alex LapanowskiJul 23, 2013 at 14:29
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