On a measurable space $(\Omega, F)$, where $\Omega$ is a set of outcomes and $F$ is a $\sigma -$field, what exactly is the domain of a probability measure $P$? If it's a specific $\sigma -$field such as $F$, then how can we define the independence on sets from multiple spaces? And it can't be the whole power set because we can construct sets that have no probability. I thought the whole point of using a $\sigma -$field was to make the domain of probability measures better to define.


So in the textbook $Probability: \ Theory \ and \ Examples$, by Rick Durrett, he says "... $\sigma -$fields $\mathcal{F}_1, \ldots , \mathcal{F}_n$ are independent if whenever $A_i \in \mathcal{F}_i$ for $i=1, \ldots, n$, we have $$P(\cap_{i=1}^{n} A_i)= \prod_{i=1}^n P(A_i)$$..."

How can $P$ be defined on all these $\sigma-$fields? I thought a probability measure was always w/respect to a given $\sigma-$field.

  • $\begingroup$ I'm afraid I don't see the question about independence and other probability spaces: what does the domain of $P$ have to do with that? Clarify a bit more? Give an example? $\endgroup$ – Henno Brandsma Feb 18 at 22:28
  • $\begingroup$ Apologies, I'll edit my question $\endgroup$ – JasonM Feb 18 at 22:40
  • $\begingroup$ Ok I edited my question $\endgroup$ – JasonM Feb 18 at 22:58
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    $\begingroup$ @JasonM the sigma fields $\mathscr{F}_i$ are contained in $\mathscr{F}$ (so $\mathscr{F}$ is at least as large as $\sigma(\cup \mathscr{F}_i)$). $\endgroup$ – Will M. Feb 18 at 23:00
  • $\begingroup$ Without more context (I don't have the book) it also seems confusing to me. One should expect that $P$ is defined on the $\sigma$-field generated by all $\mathcal{F}_i$. $\endgroup$ – Henno Brandsma Feb 18 at 23:03

The domain of $P$ is definitely $F$, the $\sigma$-field on $\Omega$ (it is in some cases the power set of $\Omega$, e.g. for discrete measures on at most countable $\Omega$).

$P$ a function from $F$ to $[0,1]$ obeying certain axioms. I don't see a relation to independence.

  • $\begingroup$ Yes this was my understanding. I hope my edit clarifies my question $\endgroup$ – JasonM Feb 18 at 22:59
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    $\begingroup$ The relation is that the sigma fields $\mathscr{F}_i$ are subset of $\mathscr{F},$ (otherwise the independence is a nonsensical issue). $\endgroup$ – Will M. Feb 18 at 23:03

Thanks to Will M. it appears the most sensible $\sigma-$field for $P$ to be defined on is a $\sigma-$field containing $\sigma(\cup_{i=1}^n \mathcal{F}_i)$. Without the experience, I had difficulty seeing this, so thanks to everyone who commented.


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