Questions: Random Variables as Measurable Functions

Question 1

Let Ω = [0,1] with the Borel sigma-algebra and Lebesgue measure. A function X: Ω → ℝ is proposed such that X⁻¹({1}) is a non-measurable subset of [0,1]. Why is X not a valid random variable?

• A. Because X can only equal 1 on a set of measure zero
• B. Because the event {ω: X(ω) = 1} is not in ℱ, so P(X = 1) is undefined
• C. Because {1} is not a Borel set in ℝ, so the preimage condition does not apply
• D. Because random variables can only map to bounded intervals, not arbitrary real values

Answer: (B)

Measurability requires X⁻¹(B) ∈ ℱ for all Borel sets B. The singleton {1} is a Borel set. If X⁻¹({1}) is non-measurable (not in ℱ), then the event 'X = 1' falls outside the sigma-algebra, and P(X = 1) is undefined — P is only defined on elements of ℱ. This is precisely why measurability is the defining condition: it guarantees every numerical statement about X translates to an event with a well-defined probability. Singletons are always Borel sets, so option C is wrong.

Question 2

A student defines Y(ω) = 1 for all ω ∈ Ω (a constant function). Is Y a valid random variable for any probability space (Ω, ℱ, P)?

• A. No — constant functions have no randomness and therefore cannot be random variables
• B. Yes — for any Borel set B, Y⁻¹(B) is either Ω or ∅, both of which are in every ℱ by definition
• C. Only if P(Ω) = 1, which requires the measure to be normalized
• D. No — its range is a single point rather than a Borel set

Answer: (B)

For any Borel set B, Y⁻¹(B) = Ω if 1 ∈ B, and ∅ if 1 ∉ B. Both Ω and ∅ belong to every sigma-algebra by definition (sigma-algebras must contain Ω and be closed under complements). Therefore Y is measurable on any probability space, making it a valid random variable. Its distribution is a point mass at 1: P(Y = 1) = 1. This shows the formal definition includes deterministic constants as a special case — 'random' in the mathematical sense does not require uncertainty.

Question 3

Nearly every function from Ω to ℝ is a random variable, as long as the probability space (Ω, ℱ, P) is well-defined.

• T. True
• F. False

Answer: False

Measurability is an additional requirement. A function X: Ω → ℝ is a random variable only if, for every Borel set B ⊆ ℝ, the preimage X⁻¹(B) is in ℱ. Non-measurable functions from Ω to ℝ exist — their construction typically involves the Axiom of Choice (e.g., Vitali sets). ℱ is generally not the power set of Ω; it is a strict subset, so some functions will map events to sets outside ℱ, making probability statements about them undefined.

Question 4

For a continuous random variable, the distribution function F(x) = P(X ≤ x) is well-defined without any need for the measurability condition.

• T. True
• F. False

Answer: False

The distribution function F(x) = P(X ≤ x) presupposes that P(X ≤ x) is defined. For P(X ≤ x) to be defined, {ω: X(ω) ≤ x} = X⁻¹((−∞, x]) must be in ℱ, since P is only defined on ℱ. The set (−∞, x] is a Borel set, so this is exactly the measurability condition. The distribution function does not stand independently of measurability — it implicitly requires it. The measure-theoretic definition is not a formality added on top of the familiar framework; it is the foundation that makes the framework rigorous.

Question 5

What is the measurability condition for a random variable, and why is it needed? Explain using the preimage concept.

Model answer: A function X: Ω → ℝ is measurable if for every Borel set B ⊆ ℝ, the preimage X⁻¹(B) = {ω ∈ Ω : X(ω) ∈ B} belongs to ℱ. It is needed because ℱ is precisely the collection of subsets of Ω that have probabilities — P is defined on ℱ and not outside it. To assign a probability to 'X ∈ B,' that event must be in ℱ, which requires X⁻¹(B) ∈ ℱ. Without measurability, X⁻¹(B) might fall outside ℱ, making P(X ∈ B) undefined. Measurability guarantees that every numerical statement about X translates to an event with a well-defined probability.

The preimage intuition: measurability means the function 'respects' the sigma-algebra structure — it pulls measurable sets in ℝ back to measurable sets in Ω. This is the bridge that connects the abstract probability space to the numerical outcomes we care about.