how to prove a field but not a sigma field

Daniel Parker

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21-01-2025

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This article explains how to demonstrate that a collection of sets forms a field (also known as an algebra of sets) but not a σ-field (σ-algebra). Understanding the distinction between these two fundamental concepts in measure theory is crucial. We'll delve into the definitions, explore the key differences, and work through examples showing how to prove one but not the other.

What is a Field (Algebra of Sets)?

A field $\mathcal{F}$ is a collection of subsets of a universal set Ω that satisfies the following three axioms:

1. Ω ∈ $\mathcal{F}$: The universal set Ω is in $\mathcal{F}$.
2. Closure under Complementation: If A ∈ $\mathcal{F}$, then A^c ∈ $\mathcal{F}$ (where A^c denotes the complement of A in Ω).
3. Closure under Finite Unions: If A, B ∈ $\mathcal{F}$, then A ∪ B ∈ $\mathcal{F}$. This extends to any finite union of sets in $\mathcal{F}$.

It's important to note that closure under finite unions implies closure under finite intersections due to De Morgan's laws.

What is a Sigma-Field (Sigma-Algebra)?

A σ-field (σ-algebra) is a more restrictive structure. It's a collection of subsets of Ω that satisfies the same three axioms as a field, plus an additional crucial axiom:

4. Closure under Countable Unions: If {A_n} is a countable sequence of sets in $\mathcal{F}$, then ∪_n=1^∞ A_n ∈ $\mathcal{F}$.

This seemingly small addition has significant consequences. A σ-field is closed under countable unions (and intersections), while a field is only closed under finite unions (and intersections).

The Key Difference: Countable vs. Finite Unions

The core difference lies in the type of unions allowed. A field only needs to be closed under finite unions, while a σ-field must also be closed under countable (potentially infinite) unions. This means any σ-field is also a field, but not all fields are σ-fields.

Proving a Field but Not a Sigma-Field: A Step-by-Step Approach

To prove a collection of sets $\mathcal{F}$ is a field but not a σ-field, follow these steps:

1. Prove it's a field: Verify that the three axioms of a field are satisfied. Show that Ω ∈ $\mathcal{F}$, that complements of sets in $\mathcal{F}$ are also in $\mathcal{F}$, and that finite unions of sets in $\mathcal{F}$ are in $\mathcal{F}$.

2. Find a counterexample to the countable union axiom: Construct a countable sequence of sets {A_n} such that each A_n ∈ $\mathcal{F}$, but ∪_n=1^∞ A_n ∉ $\mathcal{F}$. This demonstrates that $\mathcal{F}$ fails to satisfy the fourth axiom of a σ-field.

Example: The Field of Finite and Cofinite Sets

Let Ω = ℝ (the set of real numbers). Consider the collection $\mathcal{F}$ consisting of all finite subsets of ℝ and their complements (cofinite sets). Let's prove it's a field but not a σ-field.

1. Proof that $\mathcal{F}$ is a field:

   • Ω is a cofinite set (its complement is the empty set, which is finite), so Ω ∈ $\mathcal{F}$.
   • If A ∈ $\mathcal{F}$, then either A is finite or cofinite. If A is finite, A^c is cofinite and thus in $\mathcal{F}$. If A is cofinite, A^c is finite and thus in $\mathcal{F}$.
   • Let A and B be two sets in $\mathcal{F}$. If both are finite, their union is finite and thus in $\mathcal{F}$. If one is finite and the other cofinite, their union is cofinite and thus in $\mathcal{F}$. If both are cofinite, their union is also cofinite and in $\mathcal{F}$. Therefore, $\mathcal{F}$ is closed under finite unions.

2. Proof that $\mathcal{F}$ is not a σ-field: Consider the countable sequence of sets A_n = {n} for n ∈ ℕ (the set of natural numbers). Each A_n is a finite set, and therefore A_n ∈ $\mathcal{F}$. However, the countable union ∪_n=1^∞ A_n = ℕ, which is neither finite nor cofinite. Therefore, ∪_n=1^∞ A_n ∉ $\mathcal{F}$, proving that $\mathcal{F}$ is not a σ-field.

This example clearly illustrates how a collection of sets can be a field but not a σ-field. Remember, the key lies in the difference between finite and countable unions. By carefully constructing examples and applying the definitions, you can rigorously prove the nature of any given collection of sets.