0

When I typed my question the following link came up in the 'similar questions' list and I thought great, there's an answer to my question.

Why can a probability measure be defined over power set of countable sample space?

However, the answers and comments offered did not answer (or even address) @majmun's question at all, at least not from my perspective, so I'm posting my question: is there a proof that the power set of a countable sample space will support a $\textbf{probability}$ measure? Not a Dirac point measure, not a counting measure, a probability measure.

TonyK
  • 425
  • 3
    If the set is ${\mathbb N}$ you can let the measure of ${k}$ be ${1 \over 2^k}$ and then extend by countable additivity. – Zarrax May 21 '22 at 21:35
  • 2
    The Dirac measure is a probability measure. So what are you asking exactly? Do you have a clear understanding of what it means to be a probability measure? Do you want to know why there will be a probability measure with certain properties, or subject to certain conditions? If so, you should say what they are. – Izaak van Dongen May 21 '22 at 21:39
  • Perhaps you have read something along the lines of "if $\Omega$ is countable, then probability measures just correspond to mass functions, so by convention take $\mathcal F = \mathcal P(\Omega)$...". This is not the same fact as "powersets of countable sets always support some probability measure" - rather you can state this fact as "a probability measure on any $\sigma$-algebra on a countable set always extends to the powerset". This seems relevant.. – Izaak van Dongen May 21 '22 at 21:54
  • @IzaakvanDongen. Point taken on the Dirac measure. The link was interesting. – TonyK May 22 '22 at 22:08

0 Answers0