Measurable Spaces & Topological Spaces, an Analogy

While I was in Indiana a few weeks ago at Notre Dame for their Geometry & Topology conference I had a very enlightening conversation with my roommate at the time (and now good friend). He had come across a small section in one of Rudin’s analysis textbooks which highlighted an analogy between measurable spaces and topological spaces. I’d like to dive into that analogy in what follows.

Measurable Spaces & Topological Spaces

The analogy begins right with the definition of a measurable space and a topological space. So, lets restate these definitions. Given a set X, a set of subsets of X, denoted \Sigma, is called a \sigma-algebra on X if the following three properties are satisfied:

1. X \in \Sigma.

2. If A \in \Sigma then A^{c} \in \Sigma.

3. If A_{n} \in \Sigma for n = 1,2,\ldots, then \bigcup_{n = 1}^{\infty}A_{n} \in \Sigma.

In other words, \Sigma is nonempty, closed under complements, and closed under countable unions. It’s easy to check \varnothing \in \Sigma, and that \Sigma is closed under finite unions, countable intersections, and finite intersections. We call the pair (X,\Sigma) a measurable space, but we often refer to X as a measurable space without mention of the \sigma-algebra.

Given a set X, a set of subsets of X, denoted \mathcal{T}, is called a topology on X if the following three properties are satisfied:

1. X,\varnothing \in \mathcal{T}.

2. If U_{n} \in \mathcal{T} for n = 1,2,\ldots, then \bigcup_{n = 1}^{\infty}U_{n} \in \mathcal{T}.

3. If U_{i} \in \mathcal{T} for i = 1,2,\ldots,k, then \bigcap_{i = 1}^{k}U_{i} \in \mathcal{T}.

In other words, \mathcal{T} is nonempty, closed under arbitrary unions, and closed under finite intersections. We call the sets U \in \mathcal{T} the open sets of X, and we call a subset of X closed if its complement is open. The pair (X,\mathcal{T}) is called a topological space and we often refer to X as a topological space without mentioning the topology explicitly.

The similarity of these two concepts already starts to shine as a \sigma-algebra and a topology are both a collections of subsets of the ambient space that are nonempty and satisfy two closure properties. One might then ask the question if there exists a space X that is both a measurable space and a topological space such that \Sigma = \mathcal{T}. This is true as we will see.

Maps on Measurable Spaces and Topological Spaces

Mathematical spaces are important in their own right, but we often learn much more information about a space by studying the structure-preserving maps to-and-from it. For measurable spaces our maps are the measurable functions, and for topological spaces they are the continuous functions.

We say that a map f:X \to Y between two measurable spaces X and Y is measurable if the preimage f^{-1}(A) is in the \sigma-algebra for X for every set A in the \sigma-algebra for Y. Similarly, a map f:X \to Y between two topological spaces is continuous if f^{-1}(U) is an open set in X for every open set U in Y.

The reason these definitions are so similar is that the structure of measurable spaces and topological spaces is entirely dependent on a collection of subsets of the underlying space and inverse images respect arbitrary unions, arbitrary intersections, and complements.

Measurable Spaces Generated by A Topology

Here’s an easy fact: If X is a set, any collections of its subsets \mathcal{S} gives rise to a \sigma-algebra on X such that the collection of subsets is contained in the \sigma-algebra. We call this \sigma-algebra the \sigma-algebra generated by \mathcal{S}. The idea is that you take the intersection of all the \sigma-algebras on X containing \mathcal{S} (this intersection is nonempty because the set of all subsets of X is a \sigma-algebra) and prove that this is a \sigma-algebra.

Something especially interesting happens if we let X be a topological space. Indeed, if X is a topological space then the topology of X gives rise to a \sigma-algebra \mathcal{B} on X so that we may treat X as a measurable space as well. In this case, every open and closed set belongs to \mathcal{B} and we call the members of \mathcal{B} the Borel sets of X. If X is a topological space then we can treat it as a measurable space with the Borel sets being the measurable sets. In this case we call X a Borel space. If X and Y are two Borel spaces then any continuous function f:X \to Y is measurable since preimages respect arbitrary unions, arbitrary intersections, and complements.

Measures

Before we look at a couple of examples, there’s another important topic that needs some discussion, namely positive measures. Given a measurable space X a positive measure (or simply a measure) \mu on X is a function \mu:\Sigma \to [0,\infty] that is countably additive. That is, if \{A_{n}\}_{n = 1}^{\infty} is a collection of pairwise disjoint measurable sets, then

\mu\left(\displaystyle{\bigcup_{n = 1}^{\infty}A_{n}}\right) = \displaystyle{\sum_{n = 1}^{\infty}\mu(A_{n})}.

It is usually required to assume \mu(A) < \infty to weed-out odd examples. It’s easy to show that the measure \mu satisfies some basic properties such as \mu(\varnothing) = 0 (measurable sets A with \mu(A) = 0 we call sets of measure 0) and that if A and B are measurable sets with A \subseteq B then \mu(A) \le \mu(B).  Intuitively, we can think of a measure as a function which assigns a “size” to the measurable sets on the ambient space. If we have a measurable space X with a measure \mu, then X is called a measure space with measure \mu.

Examples

1. Let X be any set and let \Sigma be the power set of X. Then X is a measurable space. If S \subseteq X, define \mu(S) = |S| where |S| = \infty if S is infinite. \mu is called the counting measure on X, making X into a measure space.

2. Let X = \{0,1\} and again let \Sigma be the power set of X. We can make X into a measure space by giving it the measure \mu(\{0\}) = \mu(\{1\}) = 1/2. Since \mu is countably additive \mu(X) = 1. This is a prototypical example of a probability space (a measure space where \mu(X) = 1), and its the space used to model a fair coin flip since either the coin is heads (0) or tails (1) and each event has an equal chance of occurring (\mu(\{0\}) = \mu(\{1\}) = 1/2).

3. Let X = \mathbb{R}^{n}. Then there exists a measure \mathcal{L} called the Lebesgue measure which is translation invariant, every subset of a set of measure 0 is measurable and has measure 0 (a measure with this property is called complete). In essence, this measure encompasses the idea of measuring n-dimensional volume to a measurable subset of \mathbb{R}^{n}.

4. Let X be a locally compact Hausdorff topological group. Then there exists a measure \mu called the Haar measure which we can think of as assigning “volume” to Borel subsets of X such that \mu is translation invariant and complete. Since \mathbb{R}^{n} with the usual topology is locally compact Hausdorff it has a Haar measure, and this Haar measure is precisely the Lebesgue measure.

I hope you now enjoy the elegance of the analogy between measurable spaces and topological spaces as much as I do! That’s all for now.


References

Papa Rudin – Walter Rudin

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