Questions: Sigma-Algebras: Formal Construction

Question 1

Which of the following best explains why σ-algebras require closure under *countable* unions specifically, rather than arbitrary unions?

• A. 'Countable' is a historical convention; allowing arbitrary unions would work equally well for measure theory
• B. Restricting to countable unions prevents assembling unmeasurable sets from measurable atoms, while still supporting limits and infinite series
• C. σ-algebras were defined with countable unions because only countably many events can occur in practice
• D. Countable unions are always finite, which keeps the algebra tractable

Answer: (B)

The word 'countable' is load-bearing. Countable operations are exactly what you need for limits, infinite series, and probability over sequences of events. Allowing *arbitrary* (uncountable) unions would let you assemble non-measurable sets from measurable atoms — the same pathology that leads to paradoxes like Banach-Tarski. Option A is the most common misconception: the restriction is not a convention but a precise mathematical necessity.

Question 2

On X = {1, 2, 3}, what is σ({{1}, {2}}), the σ-algebra generated by the sets {1} and {2}?

• A. {∅, {1}, {2}, {1,2}, {1,2,3}} — just ∅, the generators, their union, and X
• B. {∅, {1}, {2}, {1,2,3}} — the generators plus the two required elements
• C. {∅, {1}, {2}, {3}, {1,2}, {1,3}, {2,3}, {1,2,3}} — the full power set
• D. {∅, {1,2,3}} — only ∅ and X are required

Answer: (C)

Starting with {1} and {2}, we must add complements: {2,3} and {1,3}. Then their union {1}∪{2} = {1,2}, and the complement of {1,2} = {3}. Once {3} is in the algebra, all remaining subsets follow. The generated σ-algebra ends up being the full power set 𝒫({1,2,3}). The common mistake is to stop after adding just the generators and their obvious unions, forgetting that complements force in much more.

Question 3

A σ-algebra on X is automatically closed under countable intersections, even though this is not listed as an axiom.

• T. True
• F. False

Answer: True

This follows from De Morgan's law: a countable intersection ∩Aₙ = complement of the countable union of complements ∪(Aₙᶜ). Since a σ-algebra is closed under complements and countable unions, countable intersections are derived for free. Students sometimes think intersection closure must be stated separately — it does not.

Question 4

The collection of most subsets of ℝ that are either finite or have finite complement (the 'co-finite' collection) is a σ-algebra on ℝ.

• T. True
• F. False

Answer: False

This collection fails closure under countable unions. Each singleton {n} for n ∈ ℕ is finite and hence in the collection. But their countable union ∪{n} = ℕ is infinite, and its complement ℝ \ ℕ is also infinite — so ℕ is not in the collection. The co-finite collection is an algebra (closed under finite unions and complements) but not a σ-algebra, which requires countable unions.

Question 5

Why must a function f: X → ℝ have the property that f⁻¹(B) ∈ 𝒜 for every Borel set B, in order to be called 'measurable'? What does this condition preserve?

Model answer: The condition ensures that the σ-algebra structure of the domain is compatible with the σ-algebra structure of the codomain. If preimages of measurable sets are measurable, then integration and probability calculations involving f are well-defined — you can assign measures to events like {x : f(x) ≤ c}. Without this condition, applying f could take measurable sets to unmeasurable ones, making integration impossible.

The measurability condition on functions mirrors the continuity condition in topology (preimages of open sets are open), but for σ-algebras. It is not about f being 'nice' in a geometric sense — it is about preserving the measurability structure that makes integration and probability coherent. A function that fails this condition could map events we can measure into events we cannot, breaking the entire framework.