Example of a category whose objects are measurable spaces and morphisms are measurable functions

In this post we consider a category whose objects are measurable spaces and morphisms are measurable functions. We limit ourselves to two objects, which let us enumerate all possible morphism. By doing so we get at better feeling of what is going on when working with categories and measurable spaces.

Let $\Omega = \{0, 1\}$ . The only $\sigma$ -algebras that can be created from this set are $\mathcal{F}_1 = \{\emptyset, \Omega\}$ and $\mathcal{F}_2 = \{\emptyset, \{0\}, \{1\}, \Omega\}$ . By paring $\Omega$ with the $\sigma$ -algebras, we get two measurable spaces. These will constitute objects in our category $\mathcal{C}$ . Further, measurable functions to and from these objects are the morphisms of $\mathcal{C}$ .

We denote the objects $a = (\Omega, \mathcal{F}_1)$ and $b = (\Omega, \mathcal{F}_2)$ and proceed by enumerating all morphisms from $a$ to $b$ , which are members of a set $\text{Hom}_{\mathcal{C}}(a,b)$ :

Let $X_1(0) = X_1(1) = 0$ . This is a measurable function from $a$ to $b$ because $X_1^{-1}(F) = \{\omega \in \Omega:\, X_1(\omega) \in F\} \in \mathcal{F}_1,\, \forall F \in \mathcal{F}_2$ . Note that we have that $X_1^{-1}(\{0\}) = \Omega \in \mathcal{F}_1$ and $X_1^{-1}(\{1\}) = \emptyset \in \mathcal{F}_1$ .
Let $X_2(0) = X_2(1) = 1$ . This is a measurable function, which is checked similarly as for $X_1$ .
Let $X_3(0) = 0$ and $X_3(1) = 1$ . This is not a measurable function because $X_3^{-1}(\{0\}) = \{0\} \notin \mathcal{F}_1$ .
Let $X_4^{-1}(0) = 1$ and $X_4(1) = 0$ . This is not a measurable function because $X_4(\{0\}) = \{1\} \notin \mathcal{F}_1$ .

Thus $\text{Hom}_\mathcal{C}(a,b) = \{X_1, X_2\}$ . Similarly, if we use $X_1, X_2, X_3, X_4$ as functions from $b$ to $b$ , we find that $\text{Hom}_\mathcal{C}(b,b) = \{X_1, X_2, X_3, X_4\}$ (these functions have the same domain and codomain as the functions from $a$ to $b$ , therefore it makes sense to ”reuse” them despite being different morphism). Note that there’s more measurable functions from $b$ to $b$ , than from $a$ to $b$ . This is because now:

$X_3^{-1}(\{0\}) = \{0\} \in \mathcal{F}_2$ and $X_3^{-1}(\{1\}) = \{1\} \in \mathcal{F}_2$ . So $X_3$ is a measurable function from $b$ to $b$ . We may verify similarly that $X_4$ is a measurable function.

Continuing in the same fashion we get $\text{Hom}_\mathcal{C}(a,a) = \{X_1, X_2, X_3, X_4\}$ and $\text{Hom}_\mathcal{C}(b,a) = \{X_1, X_2, X_3, X_4\}$ .
We further announce that $X_3$ , the identity function, is the identity morphism of $\text{Hom}_\mathcal{C}(a,a)$ as well as of $\text{Hom}_\mathcal{C}(a,a)$ , i.e. $X_3 = id_a = id_b$ . Finally, a category must have a composition formula such that for every three objects $x$ , $y$ , $z$ of C:

$\circ: \text{Hom}_\mathcal{C}(y,z) \times \text{Hom}_\mathcal{C}(x,y) \rightarrow \text{Hom}_\mathcal{C}(x,z).$

This is a function which corresponds to regular function composition. Now we may verify that that the category laws hold (Spivak, 2014, p. 204):

We must have that $f \circ id_x = f$ and $id_y \circ f = f$ for all $x, y \in \text{Ob(}\mathcal{C}\text{)}$ (right-hand-side of the expression denotes the collection of objects of $\mathcal{C}$ ) and every morphism $f: x \rightarrow y$ . Indeed, this follows from function composition and the definition of the identity function: $f \circ id_x = f(id_x(\omega)) = f(\omega),\, \omega \in \Omega$ and $id_y \circ f = id_y(f(\omega)) = f(\omega),\, \omega \in \Omega.$
We must have that $(h \circ g) \circ f = h \circ (g \circ f) \in \text{Hom}_\mathcal{C}(w,z)$ , such the that two ways to compose yield the same object in $\text{Hom}_\mathcal{C}(w,z)$ , for all $w, x, y, z \in \text{Ob(}\mathcal{C}\text{)}$ , where $f: w \rightarrow x,\, g: x \rightarrow y,\, h: y \rightarrow z$ are morphisms ( $w, x, y, z$ may not be unique, so there’s no problem that we in this case only have two objects). Indeed, this follows from function composition: $(h \circ g) \circ f = h \circ g (f(\omega)) = h(g(f(\omega))),\, \omega \in \Omega.$ and $h \circ (g \circ f) = h(g \circ f(\omega)) = h(g(f(\omega))),\, \omega \in \Omega.$