Motivation for defining $\sigma$-algebra on the codomain

Question

In probability theory, we start with a probability triple $(\Omega , \mathcal{F} , \mathbb{P})$. The purpose of defining $\mathcal{F}$ is clear to me: we postulate three intuitive axioms for $\mathbb{P}$ and, to avoid paradoxes, restrict our consideration to certain subsets, $\mathcal{F} \subset 2^\Omega$, which satisfy the properties:

1. $\Omega \in \mathcal{F},$
2. $A \in \mathcal{F} \implies A'\in \mathcal{F},$
3. $A_1,A_2,A_3,\dots \in \mathcal{F} \implies A_1 \cup A_2 \cup A_3 \cup \dots \in \mathcal{F}.$

These properties ensure that the axioms for $\mathbb{P}$ are well-defined. For example, closure under countable unions allows us to make sense of statements like $$\mathbb{P}(A_1 \cup A_2 \cup A_3 \cup \dots) = \mathbb{P}(A_1)+\mathbb{P}(A_2) + \mathbb{P}(A_3)+\dots$$ When we move to the next step and define functions from $\Omega$ to $\mathbb{R}$, it’s clear that because $\mathbb{P}$ only considers subsets in $\mathcal{F}$, we should require that for every subset $B \subset \mathbb{R}$, the preimage $f^{-1}(B) = \{\omega\in \Omega : f(\omega)\in B\}\in \mathcal{F}$ lies in $\mathcal{F}$. However, I don’t fully understand why we restrict the subsets of the codomain and, specifically, why we require the codomain to have a $\sigma$-algebra as well. Could you explain the motivation for this part of the definition of a measurable function?

Answer 1

Just for clarification: It's not true that for a random variable $f^{-1}(\mathcal{P}(\mathbb{R}) \subseteq \mathcal{F}$ holds. Such a $f$ is ofcourse a r.v. but it isn't the definition.

Coming to the question, it turns out that collection of sets on the codomain whose preimage belong to $\mathcal{F}$ is a $\sigma-$algebra.(Follows by simple set theoretic computations) (See Rudin Real and Complex Analysis Theorem 1.12 (a)) Hence, naturally there arises a sigma algebra on the codomain when we deal with a measure space.

Answer 2

With the insights provided by Nicholas Burbank's comments, I believe I’ve found a satisfying explanation.

As mentioned earlier, we define the $\sigma$-algebra $\mathcal{F}$ for the input space $\Omega$ and seek to define a mapping $X$ from the input space to the output space $\Omega'$, which is typically $\mathbb{R}$. Since our goal is to work with the measure $\mathbb{P}$, we must consider only subsets of $\Omega'$ whose preimages lie in $\mathcal{F}$. The collection of such subsets is given by: $$\begin{align*} X(\mathcal{F})=&\;\{E\subseteq \Omega'\,|\,\exists A\in \mathcal{F}:X^{-1}(E)=A\} \end{align*}.$$ Using the properties of preimages, it can be shown that $X(\mathcal{F})$ forms a $\sigma$-algebra on $\Omega'$, known as the induced $\sigma$-algebra by $X$ and $\mathcal{F}$ or pushforward $\sigma$-algebra.

1. Containment of $\Omega'$: As $X$ is a mapping, it's clear that $X^{-1}(\Omega') = \Omega\in \mathcal{F}$. Hence $\Omega' \in X(\mathcal{F}).$
2. Closure under Complements: Take $E\in X(\mathcal{F})$. So there exists $A \in \mathcal{F}$ such that $A = X^{-1}(E)$. We have $X^{-1}(\Omega' \setminus E) = X^{-1}(\Omega') \setminus X^{-1}(E) = \Omega \setminus X^{-1}(E) = \Omega \setminus A \in \mathcal{F}$. This shows that $\Omega' \setminus E\in X(\mathcal{F}).$
3. Closure under Countable Unions: Let $E_1,E_2,E_3, \dots \in X(\mathcal{F})$. So there exist $A_1,A_2,A_3,\dots \in \mathcal{F}$ such that $A_1 = X^{-1}(E_1),A_2 = X^{-1}(E_2),A_3 = X^{-1}(E_3),\dots \in \mathcal{F}$. We have $X^{-1}(E_1 \cup E_2 \cup E_3 \cup \dots ) = X^{-1}(E_1) \cup X^{-1}(E_2) \cup X^{-1}(E_3) \cup \dots = A_1 \cup A_2 \cup A_3 \cup \dots \in \mathcal{F}$, which shows $E_1 \cup E_2 \cup E_3 \cup \dots \in X(\mathcal{F})$.

Thus, it is natural to equip $\Omega'$ with the $\sigma$-algebra $X(\mathcal{F})$ and if we want to be able to deal with multiple mappings, it's natural to equip $\Omega'$ with a single $\sigma$-algebra which works for all of the mappings. In addition, we want to have as many as possible mappings and to achieve that, we should make the $\sigma$-algebra on $\Omega'$ as small as possible. See following answers for more details.

Nate Eldredge's answer:

The moral is this: To get as many $(\mathcal{B}_X,\mathcal{B}_Y)$-measurable functions $f : X \to Y$ as possible, one wants $\mathcal{B}_X$ to be as large as possible, so it makes sense to use a complete $\sigma$-algebra there. (You already know some of the nice properties of this, e.g. an a.e. limit of measurable functions is measurable.) But one wants $\mathcal{B}_Y$ to be as small as possible. When $Y$ is a topological space, we usually want to be able to compose $f$ with continuous functions $g : Y \to Y$, so $\mathcal{B}_Y$ had better contain the open sets (and hence the Borel $\sigma$-algebra), but we should stop there.

Michael Greinecker's answer:

On a more conceptual note, the less measurable sets you have in your codomain, the easier it is for a function to be measurable. And if a random variable should represent a random quantity, then all empirically interesting questions can be formulated in terms of simple intervals and their combinations. For, say, statistical applications there is no empirical difference between Borel sets and a Borel set modified by a null set. The distributions (on the reals) commonly applied can usually be given by a cumulative distribution function and such a function essentially determines the probability of intervals.

In the case where $\Omega' = \mathbb{R}$, to answer practical questions, we need all intervals $[a,b]$ with $a,b \in \mathbb{R}$ to be included in the $\sigma$-algebra. This implies that the $\sigma$-algebra must contain the $\sigma$-algebra generated by these intervals, which is the Borel $\sigma$-algebra. At this point, we can stop, as we have a suitable $\sigma$-algebra on $\mathbb{R}$. However, if we extend the $\sigma$-algebra further and make it larger, the mappings will become more restricted. For example, using the Lebesgue $\sigma$-algebra on $\mathbb{R}$ imposes additional constraints on the mappings and may lead to some unexpected results.