Let μ be Lebesgue measure on [0,1], and let ν be the measure defined by ν(A) = ∫_A x² dx. Is ν absolutely continuous with respect to μ, and what is dν/dμ?
ANo — ν is not absolutely continuous because it assigns different weights to sets than μ does
BYes — ν ≪ μ because any Lebesgue-null set also has ν-measure zero, and dν/dμ = x²
CYes — ν ≪ μ, but the Radon-Nikodym derivative cannot be identified with x² since it is not integrable on all of ℝ
DThe answer depends on whether ν is σ-finite, which is not given
Absolute continuity ν ≪ μ means: whenever μ(A) = 0, also ν(A) = 0. If a set has Lebesgue measure zero, then ∫_A x² dx = 0 as well (integrating a non-negative function over a null set gives zero). So ν ≪ μ. The function h(x) = x² satisfies ν(A) = ∫_A h dμ for all measurable A, making it the Radon-Nikodym derivative. Option A confuses 'absolutely continuous' with 'equal' — absolute continuity is about null sets, not about whether the two measures assign the same mass to every set.
Question 2 Multiple Choice
In elementary probability, if a continuous random variable X has probability density function f(x), what is f(x) in measure-theoretic terms?
AThe derivative of the CDF with respect to x, which has no direct measure-theoretic interpretation
BThe Radon-Nikodym derivative of the probability measure P with respect to Lebesgue measure λ: f = dP/dλ
CA convenient summary of P that is not formally derived from the measure structure
DThe inverse of the CDF, normalized to integrate to 1
When P is the probability measure of a continuous random variable and λ is Lebesgue measure, P(A) = ∫_A f(x) dx for all measurable A — this is exactly the Radon-Nikodym condition ν(A) = ∫_A h dμ. The density f is the Radon-Nikodym derivative f = dP/dλ. The theorem therefore is not an abstraction disconnected from elementary probability — it is the theorem that justifies the existence and uniqueness of probability density functions, and extends the concept to arbitrary measure spaces.
Question 3 True / False
If ν ≪ μ and both measures are σ-finite, the Radon-Nikodym derivative dν/dμ is unique everywhere on the space, not just μ-almost everywhere.
TTrue
FFalse
Answer: False
The Radon-Nikodym theorem guarantees uniqueness only μ-almost everywhere. Two functions h and h' satisfying ν(A) = ∫_A h dμ = ∫_A h' dμ for all measurable A must agree μ-a.e., but they can differ on a set of μ-measure zero. Since μ-null sets are invisible to all absolutely continuous measures, this is the strongest uniqueness statement available — and it is the right one for measure theory.
Question 4 True / False
If ν ≪ μ (ν absolutely continuous with respect to μ), then necessarily μ ≪ ν as well.
TTrue
FFalse
Answer: False
Absolute continuity is not symmetric. A simple example: let μ be Lebesgue measure on [0,1] and let ν(A) = ∫_A 2·1_{[0,1/2]}(x) dx — Lebesgue measure restricted to [0, 1/2] and scaled. Any Lebesgue-null set has ν-measure zero, so ν ≪ μ. But the set (1/2, 1] has μ-measure 1/2 and ν-measure 0, so μ assigns positive mass to a ν-null set, meaning μ is not absolutely continuous with respect to ν. When both ν ≪ μ and μ ≪ ν hold, the measures are called mutually absolutely continuous or equivalent.
Question 5 Short Answer
Explain in your own words why absolute continuity of ν with respect to μ is the right condition for ν to have a density with respect to μ.
Think about your answer, then reveal below.
Model answer: A density h satisfying ν(A) = ∫_A h dμ assigns zero mass to any set A that μ considers negligible (μ(A) = 0), because an integral over a null set is always zero. So if ν has a density with respect to μ, then ν must automatically assign zero mass to every μ-null set — that is, ν ≪ μ must hold. Conversely, if ν assigns positive mass to some μ-null set, no non-negative function h could produce that mass by integrating against μ, so no density can exist. Absolute continuity is therefore both necessary and sufficient for a density to exist (given σ-finiteness).
The intuition: if μ cannot 'see' a set (μ-measure zero), then integration against μ cannot accumulate mass there either. Any measure that places positive mass on μ-invisible sets is fundamentally not derivable from μ by integration — there is no way to 'create' mass on sets that μ ignores.