1

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.

William Elliot
  • 18,025
japseow
  • 321
  • 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). – freakish Sep 28 '17 at 14:09
  • 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?" – japseow Sep 28 '17 at 14:27
  • 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. – freakish Sep 28 '17 at 14:32
  • 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?". – freakish Sep 28 '17 at 14:39
  • Hmm, I do get what you mean. How about, let's look at $M = \mathbb{R}^3$, in here, not all subsets are "meaningfully" measurable (Banach-Tarski paradox). So how does sigma-algebra axioms come in to play in this situation? – japseow Sep 28 '17 at 14:51
  • @freakish The question is just about why the axioms imposed on a sigma algebra are the way they are, given that we want the elements in the sigma-algebra to be the measurable subsets of some measure. So OP wants to know how the person that first defined a sigma-algebra was led to these specific axioms, given the stated purpose of the concept. – user2520938 Sep 28 '17 at 14:56
  • @japseow they were invented because they were useful. On the one hand people wanted to use intuition (like union of two measurable sets should be measurable). On the other hand it doesn't work with all subsets on Lebesgue $\mathbb(R)$. The compromise had to be made. And $\sigma$-algebras appeared. I'm pretty sure there were many different attempts to fix this problem. But only one survived the test of time. – freakish Sep 28 '17 at 15:03
  • @user2520938 you should ask the guy who invented it. But simply stated: necessity is the mother of invention. – freakish Sep 28 '17 at 15:06
  • 1
    "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. – Did Sep 28 '17 at 15:10
  • But from references I see online, a popular idea is: "measure theory resolves the Banach-Tarski paradox without the need to cut of the axiom of choice". What does this mean? – japseow Sep 29 '17 at 09:28

1 Answers1

2

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.

Martín-Blas Pérez Pinilla
  • 42,722
  • 4
  • 51
  • 94