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Given $(\Omega,\sigma)$ a measurable space, it is not difficult to prove that if $$\forall w \in \Omega \hspace{0.4cm}\exists \hspace{0.1cm} C_w \in \sigma, w \in \cal{C}_w / \hspace{0.3cm} w\in D, D\in \sigma \Rightarrow C_w \subset D$$ and

$$|\{ \cal{C}_w / \hspace{0.2cm} w \in \Omega \}| = \aleph_0$$

then $|\sigma| = 2^{|\aleph_0|}$. Can one say something about $|\sigma|$ if the first condition is true but $|\{ \cal{C}_w / \hspace{0.2cm} w \in \Omega \}| > \aleph_0$? Are there any conditions that allow one to calculate or bound $|\sigma|$ if the two conditions above are not satisfied? (Apart from the case $|\sigma|< \infty$)

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  • $\begingroup$ Maybe I am doing a silly observation, but in the first condition one could put $C_w = \emptyset$ independently on $w$, so that $|\{ C_w : w \in \Omega \} |= 1$. Maybe are you assuming that $w \in C_w$? $\endgroup$ – Crostul Nov 13 '14 at 0:28
  • $\begingroup$ You are right, I forgot writing it. Thank you. $\endgroup$ – somebody Nov 13 '14 at 0:37
  • $\begingroup$ This means simply that $C_w = \bigcap_{w \in D \in \sigma} D$. $\endgroup$ – Crostul Nov 13 '14 at 0:38
  • $\begingroup$ It is other way to define the same concept, yes. $\endgroup$ – somebody Nov 13 '14 at 0:42
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For all $w \in \Omega$ we have that, assumed it exists (and we are assuming it), $$C_w = \bigcap_{w \in D \in \sigma} D$$ i.e. $C_w$ is the smallest measurable set containing $w$.

It's easy to see that $\Omega' = \{ C_w : w \in \Omega\}$ is a partition of $\Omega$, so it induces an equivalence relation $v \sim w \Leftrightarrow C_v = C_w$, and $\Omega' = \Omega / \sim$.

Now, define on $\Omega'$ the following $\sigma$-algebra: $$\sigma' = \{ \{ C_w :w \in D \} : D \in \sigma\}$$

It is easy to see that the following hold:

  1. $\sigma'$ is a $\sigma$-algebra on $\Omega'$
  2. There exists an injection $\Omega' \longrightarrow \sigma'$ given by $$X \mapsto \{ X \} $$
  3. There exists a bijection $\sigma' \longrightarrow \sigma$ given by $$\{ C_w :w \in D \} \mapsto D$$

So, finally I can answer to the question. If we know $|\Omega'| > \aleph_0$, what can we say about $|\sigma|=|\sigma'|$? The answer is: all we can say is that $|\Omega'| \leq |\sigma'| \leq 2^{|\Omega'|}$, by point 2.

However, we cannot say anything more, because the $\sigma$-algebra generated by the singletons of $\Omega'$ is $$\Sigma = \{ E \subseteq \Omega' : \mbox { either $E$ is countable or its complement is countable} \}$$

which has the same cardinality of $\Omega'$ if and only if $\Omega'$ is not countable. So, we know that $\sigma' \supseteq \Sigma$, but we could have that $\sigma'$ is much bigger (e.g. $\sigma'$ could be the whole set $\mathcal{P}(\Omega')$).

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