Measurable Spaces, Measure Spaces, and Measurable Functions
We’ve spent a fair amount of time discussing rings and -rings of sets, and measures as functions on such collections. Now we start considering how these sorts of constructions relate to each other.
A “measurable space” is some set and a choice of a -ring of subsets of . We call the members of the “measurable sets” of the measurable space. This is not to insinuate that is the collection of sets measurable by some outer measure , nor even that we can define a nontrivial measure on in the first place. Normally we just call the measurable space by the same name as the underlying set and omit explicit mention of .
Since it would be sort of silly to have points that can’t be discussed, we add the assumption that every point of is in some measurable set. Commonly, it’s the case that itself is measurable — — but we won’t actually require that be a -algebra.
A “measure space” is a measurable space along with a choice of a measure on the -ring . As before, we will usually call a measure space by the same name as the underlying set and omit explicit mention of the -ring and measure. Measure spaces inherit adjectives from their measures; a measure space is called finite, or -finite, or complete if its measure is finite, -finite, or complete, respectively.
Given a measure space , we will routinely use without comment the associated outer measure and inner measure on the hereditary -ring .
As an underlying set equipped with a particular collection of “special” subsets, a measurable space should remind us of a topological space, and like topological spaces they form a category. Remember that our original definition of a continuous function: given topological spaces and , a function is continuous if the preimage of any open set is open — for any we have .
We define a “measurable function” similarly: given measurable spaces and , a function is measurable if the preimage of any measurable set is measurable — for any we have . It’s straightforward to verify that the collection of measurable spaces and measurable functions forms a category. We will set this category in its full generality aside for the moment, as is the usual practice in measure theory, but we will refer to it if appropriate to illuminate a point.
Before I close, though, I’d like to put out a question that I don’t know the answer to, and which some friends haven’t really been able to answer when I mused about it in front of them. When we dealt with topology, we were able to recast the basic foundations in terms of nets. That is, a function is continuous if and only if it “preserves limits of convergent nets” — if it sends any convergent net in the domain to another convergent net in the range, and the action of the function commutes with passage to the limit. I like this because the idea of “preserving” some structure (albeit an infinite and often-messy one) feels more natural and algebraic that the idea of inspecting preimages of open sets. And so I put the question out to the audience: what is “preserved” by a measurable function, in the same way that continuous functions preserve limits of convergent nets?
is it enough [tex] mathcal{S} [/tex] be a [tex] \sigma ring [/tex] a probability book i’m studying says the space needs a [tex] \sigma – Field[/tex]
Could you clarify please
thanks
Comment by cappa | April 27, 2010 |
A -field is a different name for a -algebra. As I point out above, in many cases actually is a -algebra (or -field). In this case, itself is in , and so it’s clear that every point in is in some measurable set.
However, we’re going to allow to just be a -ring — itself might not be in — so long as every point in is still in some measurable set.
Comment by John Armstrong | April 27, 2010 |
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Is every measurable space is topological space
Comment by M. A. Hossain | September 27, 2011 |
Not in any natural way. The obvious thing would be to try to use the -algebra of measurable sets as the collection of open sets in the topology, but it’s not necessarily closed under arbitrary unions.
Comment by John Armstrong | September 27, 2011 |
Actually my query is as follows:
If T is a sigma-algebra In X, is T also a topology in X?
Comment by M. A. Hossain | September 28, 2011 |
No, as I said; is not necessarily closed under arbitrary (uncountable) unions.
Comment by John Armstrong | September 28, 2011 |