# The Unapologetic Mathematician

## 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?

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

## 14 Comments »

1. is it enough $$mathcal{S}$$ be a $$\sigma ring$$ a probability book i’m studying says the space needs a $$\sigma – Field$$
Could you clarify please

thanks

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