The Unapologetic Mathematician

Mathematics for the interested outsider

Measurable Spaces, Measure Spaces, and Measurable Functions

We’ve spent a fair amount of time discussing rings and \sigma-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 X and a choice of a \sigma-ring \mathcal{S} of subsets of X. We call the members of \mathcal{S} the “measurable sets” of the measurable space. This is not to insinuate that \mathcal{S} is the collection of sets measurable by some outer measure \mu^*, nor even that we can define a nontrivial measure on \mathcal{S} in the first place. Normally we just call the measurable space by the same name as the underlying set X and omit explicit mention of \mathcal{S}.

Since it would be sort of silly to have points that can’t be discussed, we add the assumption that every point of X is in some measurable set. Commonly, it’s the case that X itself is measurable — X\in\mathcal{S} — but we won’t actually require that \mathcal{S} be a \sigma-algebra.

A “measure space” is a measurable space along with a choice of a measure \mu on the \sigma-ring \mathcal{S}. As before, we will usually call a measure space by the same name as the underlying set X and omit explicit mention of the \sigma-ring and measure. Measure spaces inherit adjectives from their measures; a measure space is called finite, or \sigma-finite, or complete if its measure \mu is finite, \sigma-finite, or complete, respectively.

Given a measure space (X,\mathcal{S},\mu), we will routinely use without comment the associated outer measure \mu^* and inner measure \mu_* on the hereditary \sigma-ring \mathcal{H}(\mathcal{S}).

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 (X_1,\mathcal{T}_1) and (X_2,\mathcal{T}_2), a function f:X_1\to X_2 is continuous if the preimage of any open set is open — for any O\in\mathcal{T}_2 we have f^{-1}(O)\in\mathcal{T}_1.

We define a “measurable function” similarly: given measurable spaces (X_1,\mathcal{S}_1) and (X_2,\mathcal{S}_2), a function f:X_1\to X_2 is measurable if the preimage of any measurable set is measurable — for any S\in\mathcal{S}_2 we have f^{-1}(S)\in\mathcal{S}_1. 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?

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April 26, 2010 - Posted by | Analysis, Measure Theory


  1. 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


    Comment by cappa | April 27, 2010 | Reply

  2. A \sigma-field is a different name for a \sigma-algebra. As I point out above, in many cases \mathcal{S} actually is a \sigma-algebra (or -field). In this case, X itself is in \mathcal{S}, and so it’s clear that every point in X is in some measurable set.

    However, we’re going to allow \mathcal{S} to just be a \sigma-ring — X itself might not be in \mathcal{S} — so long as every point in X is still in some measurable set.

    Comment by John Armstrong | April 27, 2010 | Reply

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  11. Is every measurable space is topological space

    Comment by M. A. Hossain | September 27, 2011 | Reply

  12. Not in any natural way. The obvious thing would be to try to use the \sigma-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 | Reply

  13. 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 | Reply

  14. No, as I said; T is not necessarily closed under arbitrary (uncountable) unions.

    Comment by John Armstrong | September 28, 2011 | Reply

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