AFAIK, sigma-algebras are defined because it there is no meaningful way one could assign measures to all subsets.

So, given the measure space $(M, \sigma, \mu)$: How are the axioms of $\sigma$ (closed under complements, countable unions and intersections) related to $\mu$ being a "meaningful" measure?

Edit: Sorry if it's a dumb question, I'm very new to this stuff.

  • $\begingroup$ What do you mean by "meaningful". There are many possible measures on a given $\sigma$-algebra and on every $\sigma$-algebra there's at least one measure, i.e. the counting measure (BTW it can be assigned to the $\sigma$-algebra of all subsets). $\endgroup$
    – freakish
    Sep 28 '17 at 14:09
  • $\begingroup$ Well, we can omit that part and just ask the question: "How does closure under complements, countable unions, etc. associate with the concept of measurements / measurable space?" $\endgroup$
    – japseow
    Sep 28 '17 at 14:27
  • $\begingroup$ I still don't understand the question. If you have a group $(G, \circ)$ then do you ask what properties of the underlying set $G$ are related to $\circ$? Or any other structure built on some structure? The reason we use measures on $\sigma$-algebras is because these properties/definitions were extracted from concrete objects we are dealing with every day. And as you said it was necessary because some concrete measure (i.e. Lebesgue measure) doesn't work on all sets. And thus a better definition was invented. $\endgroup$
    – freakish
    Sep 28 '17 at 14:32
  • $\begingroup$ But local properties of the Lebesgue measure does not necessarily generalize to all measures. In particular as I said every $\sigma$-algebra (including all subsets) has a measure. So you would have to be more specific, something like "I would like a measure with property X, is that possible?". $\endgroup$
    – freakish
    Sep 28 '17 at 14:39
  • 1
    $\begingroup$ "sigma-algebras are defined because it there is no meaningful way one could assign measures to all subsets" Although common, this is a glaring misconception. $\endgroup$
    – Did
    Sep 28 '17 at 15:10

Two intuitive reasons:

If you want countable additivity, is reasonably required that the countable union of measurable sets be measurable.

Particular case of additivity: if you want that $\mu(M) = \mu(A) + \mu(M\setminus A)$, is reasonably required that a set and its complementary be simultaneously measurable or non-measurable.

  • $\begingroup$ Ahh! I see, this does explain the axioms a little bit. $\endgroup$
    – japseow
    Sep 29 '17 at 9:26
  • $\begingroup$ @japseow, if you find the answer useful, you can accept it. $\endgroup$ Sep 29 '17 at 10:15
  • $\begingroup$ @Martín-BlasPérezPinilla But why is it important to have $\mu(M)$ defined? If not, then we don’t need the complementarity property of $\sigma$-algebras. $\endgroup$ Mar 12 '19 at 4:03
  • $\begingroup$ @YatharthAgarwal, well, the measure of the whole space surely is interesting... $\endgroup$ Mar 12 '19 at 8:41
  • 1
    $\begingroup$ @YatharthAgarwal, if you want do integrals in the whole space... Famous example: the Gaussian integral and generalizations: math.stackexchange.com/questions/126227/…. $\endgroup$ Mar 12 '19 at 9:25

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.