## 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