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In the Wikipedia article it says:

the collection of those subsets for which a given measure is defined is necessarily a $\sigma$-algebra.

Fine, but is the opposite true? Do we know for sure that all sets of sigma algebra are measurable? If the answer is no, then is it the reason why Borel sigma algebra is so widely used in probability theory?

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  • $\begingroup$ The Borel sigma algebra is useful when discussing measurable functions, since we would like continuous functions to be measurable. As for your first question, how do you define a measurable set? The definition implies the existence of a $\sigma$-algebra. $\endgroup$ – Prahlad Vaidyanathan Dec 1 '15 at 10:38
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    $\begingroup$ Your question is not clear: what $\sigma$-algebra and what measure are you talking about? $\endgroup$ – Crostul Dec 1 '15 at 10:40
  • $\begingroup$ @Crostul I'm talking about any sigma algebra - whether it's true that all elements (i.e. sets) of any sigma algebra are measurable. $\endgroup$ – user216094 Dec 1 '15 at 10:42
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    $\begingroup$ A set $A$ is not loosely measurable. It is measurable with respect to some $\sigma$-algebra. When is this the case? Exactly when $A$ is an element of that $\sigma$-algebra. $\endgroup$ – drhab Dec 1 '15 at 10:44
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    $\begingroup$ The statement is indeed meaningless if context lacks. Every outer measure induces a $\sigma$-algebra (see the answer of @Etienne) and the restriction of the outer measure to this $\sigma$-algebra is a measure. In special case a set belongs to that $\sigma$-algebra if it has equal outer an inner measure. Proving that a set is not measurable (requiring a lot of formalism) is the same as proving that it is not an element of the $\sigma$-algebra that is involved. Most cases deal with Borel- or Lebesgue $\sigma$-algebra. $\endgroup$ – drhab Dec 1 '15 at 13:42
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Here is one answer, which may answer the question as I understand it (but perhaps I do not understand the question correctly).

Start with a measure $\mu$ defined on a $\sigma$-algebra $\mathcal A$. Then define the outer measure $\mu^*$ associated with $\mu$, in the sense of Caratheodory. Once you have this outer measure, you can define the class $\mathcal A(\mu^*)$ of all $\mu^*$-measurable sets (still in the sense of Caratheodory).

Then, it is part of the Caratheodory extension theorem that $\mathcal A\subseteq \mathcal A(\mu^*)$. So, in this sense, the answer to the question is "Yes".

On the other hand, it is not necessarily tru that all measurable sets belong to the original $\sigma$-algebra $\mathcal A$...

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