Which condition defines a set E ⊆ ℝ as Carathéodory measurable?
AE is open, or can be written as a countable union of open intervals
BThe outer measure μ*(E) equals the infimum of the lengths of all open covers of E
CFor every set A ⊆ ℝ, μ*(A) = μ*(A ∩ E) + μ*(A ∩ Eᶜ)
DE belongs to the Borel σ-algebra generated by open intervals
The Carathéodory condition says E is a 'clean' partition tool — using E to split any set A neither loses nor double-counts measure. This is more general than being open or being Borel: every Borel set satisfies it, but the Lebesgue σ-algebra also includes all subsets of null sets (sets of measure zero), making Lebesgue measure complete. Options A and D describe smaller classes; option B describes how outer measure is defined, not how measurability is characterized.
Question 2 Multiple Choice
The rational numbers ℚ are dense in ℝ — between any two real numbers there is a rational. What is the Lebesgue measure of ℚ?
APositive, since ℚ is dense and 'fills' the real line in a topological sense
BUndefined, because ℚ is not a Borel set
C0 — the rationals form a countable null set
DEqual to the measure of ℝ, since ℚ intersects every open interval
ℚ is countable, so enumerate it as q₁, q₂, …. For any ε > 0, cover qₙ with an open interval of length ε/2ⁿ. The total length is ε·Σ(1/2ⁿ) = ε. Since ε is arbitrary, μ(ℚ) = 0. Density is a topological concept entirely separate from measure — a set can be topologically dense (intersecting every open interval) while having zero measure. This separation is one of the sharpest breaks between Lebesgue theory and older intuitions.
Question 3 True / False
The Lebesgue σ-algebra of measurable sets is strictly larger than the Borel σ-algebra — it includes all subsets of sets with Lebesgue measure zero.
TTrue
FFalse
Answer: True
This property is called completeness of Lebesgue measure. If a set N has measure zero, the Carathéodory condition is easily verified for any subset S ⊆ N (since μ*(A ∩ S) ≤ μ*(N) = 0 for all A). Borel sets do not include all such subsets. Completeness matters for integration: you want to be able to modify a function on a null set without changing its integrability or integral value.
Question 4 True / False
Most subset of ℝ is Lebesgue measurable.
TTrue
FFalse
Answer: False
Non-measurable sets exist — the classic example is the Vitali set, constructed using the axiom of choice by selecting one representative from each equivalence class of reals modulo the rationals. Such sets violate the Carathéodory condition. Non-measurability is not a defect of the theory; it is precisely why the theory restricts to measurable sets. The existence of non-measurable sets shows that not every set has a well-defined size consistent with countable additivity and translation invariance.
Question 5 Short Answer
Why does the existence of null sets like ℚ matter for the Lebesgue integral, and how does this distinguish Lebesgue integration from Riemann integration?
Think about your answer, then reveal below.
Model answer: The Lebesgue integral ignores behavior on sets of measure zero — a function can be changed on a null set without affecting its integral ('almost everywhere' equality). This means functions like the Dirichlet function (1 on ℚ, 0 on irrationals) are Lebesgue integrable (integral = 0, since ℚ has measure zero) but not Riemann integrable (every Riemann sum partition mixes rationals and irrationals, so upper and lower sums never agree). Lebesgue's framework handles a vastly richer class of functions by treating measure-zero sets as negligible.
The 'almost everywhere' concept — equality, continuity, or convergence holding except on a null set — is fundamental to modern analysis. Null sets are the Lebesgue integral's way of saying 'this doesn't matter for integration purposes.' The Riemann integral has no analogous concept, which is why it fails on even simple functions that differ on countable sets.