1
$\begingroup$

Let $\Omega$ a non empty set and consider and $\cal{A}$ a sigma algebra in $\Omega$. Is there always a set $\cal{C}\subset\mathscr{P}(\Omega)$, $\cal{C}$ strictly contained in $\cal{A}$ such that $\cal{A}$ is generated by $\cal{C}?$

For example, is the Lebesgue measure generated by some $\cal{C}?$

Obs. https://mathoverflow.net/questions/87838/is-every-sigma-algebra-the-borel-algebra-of-a-topology

I have already seen the discussion in the above link, but I'm not asking noting special about the class $\cal{C}$.

Also in the above case the answer is quite sophisticated. I'm wondering if in the question I ask the answer is more simple.

$\endgroup$
  • 1
    $\begingroup$ You had better put some condition on $\mathcal C$, otherwise just take $\mathcal A = \mathcal C$. $\endgroup$ – treble May 20 '18 at 19:47
  • $\begingroup$ Thanks for the warning. When I ask the question I have the condition ${\cal C}$ strictly contained in ${\cal A}$ in mind $\endgroup$ – Eduardo May 20 '18 at 19:55
1
$\begingroup$

If you take $\mathcal{C}=\mathcal A$, then you'll have a set that generates $\mathcal A$. If you want a smaller set, take $\mathcal{C}=\mathcal{A}\setminus\{\emptyset\}$.

| cite | improve this answer | |
$\endgroup$

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.