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Measurable/Continuous Functions, Initial-Product \(\sigma\)-Algebra/Topology

The following subsection deals with the consequence of the properties of preimages. Generically, it works as follows. Let \(f\colon X\to Y\) be a function and \(\mathcal{C}\) be a collection of subsets of \(Y\) that is stable under intersection, union, or complement. Then, by Pre-image Proposition, the collection \(\{f^{-1}(B)\colon B\in \mathcal{C}\}\) inherits the same stability properties. This is why measurability, as well as continuity, is defined in terms of preimages.

For ease of notation, given a collection \(\mathcal{C}\) of subsets of \(Y\), we use the notation

\[ \begin{equation} f^{-1}(\mathcal{C})=\{f^{-1}(B)\colon B\in \mathcal{C}\}. \end{equation} \]

Also, in probability theory, we use the following shorthand notation

\[ \begin{equation} \left\{ f \in A \right\}:=f^{-1}(A)=\left\{ x\in X\colon f(x)\in A \right\}. \end{equation} \]

If \(Y=\mathbb{R}\), we also use the notations

\[ \begin{equation} \left\{ f\leq t \right\}:=f^{-1}\left( (-\infty,t] \right), \quad \left\{ s < f \leq t \right\}:=f^{-1}\left((s,t]\right), \quad \text{etc.} \end{equation} \]

Definition: Measurability/Continuity

Let \((X,\mathcal{F})\) and \((Y, \mathcal{G})\) be two measurable spaces. We say that a function \(f:X\to Y\) is \(\mathcal{F}\)-\(\mathcal{G}\)-measurable if

\[ \begin{equation} f^{-1}(B) \in \mathcal{F}, \quad\text{for every measurable set }B \in \mathcal{G}. \end{equation} \]

In other words, if and only if \(f^{-1}(\mathcal{G})\subseteq \mathcal{F}\).

Let \((X,\mathfrak{S})\) and \((Y,\mathfrak{T})\) be two topological spaces. We say that a function \(f:X\to Y\) is \(\mathfrak{S}\)-\(\mathfrak{T}\)-continuous if

\[ \begin{equation} f^{-1}(O) \in \mathfrak{S}, \quad\text{for every open set } O\in \mathfrak{T}. \end{equation} \]

In other words, if and only if \(f^{-1}(\mathfrak{T})\subseteq \mathfrak{S}\).

Lemma

The composition of measurable/continuous functions is measurable/continuous, respectively.

Proof

We only prove that the composition of measurable functions is measurable, the continuous case being similar. Let \((X,\mathcal{F})\), \((Y,\mathcal{G})\), and \((Z,\mathcal{H})\) be three measurable spaces and \(f\colon X\to Y\) and \(g\colon Y\to Z\) be two measurable functions and denote \(h=g\circ f\). For every \(C\in \mathcal{H}\), it holds that

\[ \begin{equation} h^{-1}(C)=f^{-1}(g^{-1}(C))=f^{-1}(B) \quad \text{where } B=g^{-1}(C). \end{equation} \]

Since \(g\) is measurable, it follows that \(B=g^{-1}(C)\in \mathcal{G}\). Further, the measurability of \(f\) implies that \(h^{-1}(C)=f^{-1}(B)\in \mathcal{F}\) showing that \(h\) is measurable.

Lemma

Let \(f \colon X\to Y\) be a function.

  • If \((Y,\mathcal{G})\) is a measurable space, then \(f^{-1}(\mathcal{G})=\{f^{-1}(B)\colon B \in \mathcal{G}\}\) is a \(\sigma\)-algebra called the \(\sigma\)-algebra generated by \(f\) and denoted by \(\sigma(f)\).
  • If \((Y,\mathfrak{T})\) is a topological space, then \(f^{-1}(\mathfrak{T})=\{f^{-1}(O)\colon O\in \mathfrak{T}\}\) is a topology called the topology generated by \(f\) and denoted by \(\mathfrak{t}(f)\).
Proof

The proof is a straightforward application of Pre-image Proposition.

If \(f\colon X\to Y\) is an \(\mathcal{F}\)-\(\mathcal{G}\)-measurable function where \(\mathcal{F}\) is a \(\sigma\)-algebra on \(X\), then it follows that \(\sigma(f)\subseteq \mathcal{F}\). In other words, \(\sigma(f)\) is the smallest \(\sigma\)-algebra for which \(f\) is measurable. The same remark holds for continuous functions: \(\mathfrak{t}(f)\) is the smallest topology for which \(f\) is continuous. More generally, let \((f_i)\) be a family of functions \(f_i\colon X \to Y_i\) where \((Y_i,\mathcal{G}_i)\) is a family of measurable spaces. Then,

\[ \begin{equation} \sigma(f_i\colon i)=\sigma(\{f_i^{-1}(B)\colon B\in \mathcal{G}_i, i\}) \end{equation} \]

is the smallest \(\sigma\)-algebra such that each \(f_i\) is measurable. Similarly, when replacing \(\sigma\)-algebra with topologies, the same holds. These are called the initial \(\sigma\)-algebra or topology.

Definition: Product \(\sigma\)-algebra

Let \((X_i,\mathcal{F}_i)\) be a non-empty family of measurable spaces. The product \(\sigma\)-algebra, denoted by \(\otimes \mathcal{F}_i\) on the product state space \(X =\prod X_i\), is defined as the \(\sigma\)-algebra generated by the family of projections:

\[ \begin{align} \pi_i\colon X &\longrightarrow X_i, \\ x=(x_i)&\longmapsto x_i. \end{align} \]

That is,

\[ \begin{equation} \otimes \mathcal{F}_i =\sigma\left(\left\{ \pi^{-1}_i(A_i)\colon A_i\in \mathcal{F}_i, \text{ for any }i \right\}\right). \end{equation} \]

Exercice

In two dimensions, let \((X,\mathcal{F})\) and \((Y,\mathcal{G})\) be two measurable spaces, and define the projections:

\[ \begin{align} \pi_1 \colon X\times Y &\longrightarrow X, & \pi_2\colon X\times Y & \longrightarrow Y, \\ (x,y)&\longmapsto \pi_1((x,y))=x, & (x,y)&\longmapsto \pi_2((x,y))=y. \end{align} \]

In particular, for \(A\) in \(\mathcal{F}\) and \(B\) in \(\mathcal{G}\), one has:

\[ \begin{equation} \begin{split} \pi^{-1}_1(A)&=\left\{ (x,y)\colon \pi_1((x,y))=x \in A \right\}=A\times Y, \\ \pi^{-1}_2(B)&=\left\{ (x,y)\colon \pi_2((x,y))=y \in B \right\}=X\times B. \end{split} \end{equation} \]

It follows that:

\[ \begin{equation} \mathcal{F}\otimes \mathcal{G}=\sigma\left(\left\{ A\times Y, X\times B\colon A \in \mathcal{F}, B \in \mathcal{G} \right\}\right). \end{equation} \]

As an exercise, show that:

\[ \begin{equation} \mathcal{F}\otimes \mathcal{G}=\sigma\left(\left\{A\times B\colon A\in \mathcal{F}, B\in \mathcal{G}\right\}\right). \end{equation} \]

Topologies and \(\sigma\)-algebras are often complex abstract systems of sets. In practical terms, checking continuity or measurability for a function might, at first sight, seem like an impossible task. However, topologies and \(\sigma\)-algebras are often described in terms of a simple collection of sets from which they are generated. The morphism properties of pre-images allows us to restrict the verification of continuity or measurability to this simpler generating set.

Proposition

Let \(f:X\to Y\) be a function and \(\mathcal{C}\), \(\mathfrak{C}\) be collections of subsets of \(Y\). It follows that

\[ \begin{equation} \sigma(f)=f^{-1}\left(\sigma(\mathcal{C})\right)=\sigma\left( f^{-1}(\mathcal{C})\right)=\sigma\left(\left\{ f^{-1}(B)\colon B \in \mathcal{C} \right\} \right). \end{equation} \]

and

\[ \begin{equation} \mathfrak{t}(f)=f^{-1}\left(\mathfrak{t}(\mathfrak{C})\right)=\mathfrak{t}\left( f^{-1}(\mathfrak{C})\right)=\mathfrak{t}\left(\left\{ f^{-1}(B)\colon B \in \mathfrak{C} \right\} \right). \end{equation} \]

Proof

Left as an exercise.

Remark

The use of this proposition is often as follows. Let \(f:X\to Y\) be a function where \((X,\mathcal{F})\) and \((Y,\mathcal{G})\) are measurable spaces. Suppose that \(\mathcal{G}=\sigma(\mathcal{C})\) for some collection \(\mathcal{C}\) of subsets of \(Y\). It follows that \(f\) is measurable if and only if

\[ \begin{equation} f^{-1}(B)\in \mathcal{F}, \quad \text{ for every }B\in \mathcal{C}. \end{equation} \]

In particular, if \(Y=\mathbb{R}\) and \(\mathcal{G}=\mathcal{B}(\mathbb{R})\) is the Borel \(\sigma\)-algebra of \(\mathbb{R}\), then \(f\) is measurable if and only if

\[ \begin{equation} \{f\leq t\} \in \mathcal{F} \end{equation} \]

for every \(t \in \mathbb{R}\).

Corollary

Let \((X,\mathcal{F})\) be a measurable space and \((Y,\mathfrak{T})\) be a topological space. A function \(f \colon X \to Y\) is measurable with respect to the \(\sigma\)-algebra if \(f^{-1}(O)\) is measurable for any open set \(O\).

Furthermore, if the topology is generated by a countable basis \(\mathfrak{B}\), we can consider only those \(O\) in the topological basis.

In particular if \(X\) and \(Y\) are topological spaces endowed with their respective \(\sigma\)-algebra, if \(f\) is continuous then it is measurable.

Finally, if \(Y = \mathbb{R}\) and \(f\) is lower/upper semi-continuous if follows that \(f\) is measurable.

Proof

The first assertion is a direct consequence of the previous results, the seond one follows from the fact that every open set \(O\) can be written as a countable union of elements in \(\mathfrak{B}\).

As for the third one, it follows from \(f^{-1}(O)\) is open henceforth measurable for any open set \(O\).

As for the last one it follows from \(f^{-1}(]-\infty, t])\) is closed for any \(t\) henceforth measurable.

Proposition

Let \(f,g, f_n\colon X\to \mathbb{R}\) be measurable functions where \((X,\mathcal{F})\) is a measurable space. It holds that:

  • \(af+bg\) is measurable for every \(a,b \in \mathbb{R}\).
  • \(fg\) is measurable.
  • \(\max(f,g)\) and \(\min(f,g)\) are measurable.

  • \(\sup f_n\) and \(\inf g_n\) are extended real-valued measurable functions.(1)

    1. With respect to the Borel \(\sigma\)-algebra on \([-\infty,\infty]\) generated by the metric \(d(x,y)=|\arctan(x)-\arctan(y)|\), which coincides with the Euclidean topology on \(\mathbb{R}\).

  • \(\liminf f_n:=\inf_n\sup_{k\geq n}f_k\) and \(\limsup f_n:=\inf_n\sup_{k\geq n}f_k\) are extended real-valued measurable functions.

  • \(A:=\{\lim f_n \text{ exists}\}:=\{\omega \colon \lim f_n(\omega)\text{ exists}\}=\{\liminf f_n=\limsup f_n\}\) is measurable.

Proof

First, let \(\varphi:\mathbb{R}\times \mathbb{R}\to \mathbb{R}\) be a continuous function. It follows that the function \(\varphi(f,g)\) is measurable for the following reason. First, the mapping \(T:x \to \mathbb{R}\times \mathbb{R}\), \(x \mapsto (f(x),g(x))\) is measurable with respect to the product Borel \(\sigma\)-algebra on \(\mathbb{R}\times \mathbb{R}\). Indeed, for every two Borel sets \(A, B\) of the real line, it follows that

\[ T^{-1}(A\times B)=\{f\in A\}\cap\{g\in B\} \]

which is an element in \(\mathcal{F}\) by the measurability of \(f\) and \(g\). Since the product sets \(A\times B\) for \(A,B\) Borel sets in \(\mathbb{R}\) generate the Borel product \(\sigma\)-algebra on \(\mathbb{R}^2\), it follows that \(T\) is measurable. By continuity of \(\varphi\), it follows that \(\varphi\) is measurable, and therefore \(\varphi\circ T\) is measurable. Taking \(\varphi(x,y)=ax+by\), \(\varphi(x,y)=xy\), \(\varphi(x,y)=\max(x,y)\), and \(\varphi(x,y)=\min(x,y)\), the first three points follow.

Let \(a \in \mathbb{R}\). It holds that

\[ \{\sup_n f_n \leq a\}=\{f_n\leq a\colon \text{ for every }n\}=\cap_n \{f_n\leq a\} \]

which is measurable since \(\{f_n\leq a\}\) is measurable. Since \((-\infty,a]\) generates the Borel \(\sigma\)-algebra, it follows that \(\sup_n f_n\) is measurable. The same argument applies to \(\inf f_n\) using \(\{\inf f_n\geq a\}\).

The measurability of \(\liminf f_n\) follows by the same argument using countable intersection and union. The last point follows directly.