1
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ Any uncountable set is the union of its points... $\endgroup$ – Georges Elencwajg Jan 20 '14 at 10:50
2
$\begingroup$

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 $0$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.