As the title suggests, I wonder whether continuity of measures holds for uncountable operations? I.e., is it true that $E_\alpha \uparrow E \Rightarrow \lim_{\alpha}\mu(E_\alpha )= \mu (\cup_\alpha E_\alpha)$, $\alpha \in I$ for some uncountable index set $I$?

The only proofs I've seen for continuity of measure use countable additivity, and it's clear to me that uncountable additivity of measures does not make sense. However, is there a straight forward way to prove or disprove uncountable continuity?

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    $\begingroup$ Any uncountable set is the union of its points... $\endgroup$ – Georges Elencwajg Jan 20 '14 at 10:50

Not in general, no. Take $I=[0,1]$ (note the closedness at $1$!) and define $E_\alpha = \{0\}$ if $\alpha<1$ and $E_1 = [0,1]$. You now have $$\bigcup E_{\alpha} = [0,1]$$ you have $$E_\alpha \subseteq E_\beta$$ for $\alpha < \beta$, there's even a limit $\lim_{\alpha \rightarrow 1} \mu(\alpha)$, but it equals $0$.

  • $\begingroup$ Thanks,that makes it perfectly clear. $\endgroup$ – ekvall Jan 20 '14 at 11:52

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