A sequence of measurable functions f₁, f₂, f₃, … on a measure space converges pointwise to a function f everywhere. Which of the following is true about f?
Af may not be measurable, since pointwise limits do not generally preserve measurability
Bf is measurable, and this closure under pointwise limits is a key advantage over Riemann-integrable functions
Cf is measurable only if the convergence is also uniform
Df is measurable only if each fₙ is also continuous
Measurable functions are closed under pointwise limits — a property Riemann-integrable functions do NOT share. This is what gives the Lebesgue integral its power: theorems like the Dominated Convergence Theorem allow you to pass limits through integrals because the limit function is still measurable. Uniform convergence is not required; pointwise convergence suffices.
Question 2 Multiple Choice
To verify that f: (X, 𝒜) → ℝ is measurable, a student checks only whether {x : f(x) < a} ∈ 𝒜 for every real number a. Her professor says this is sufficient. Why?
AIt is not sufficient — one must also check preimages of closed sets and compact sets independently
BIt is sufficient because the half-infinite intervals (-∞, a) generate the Borel σ-algebra on ℝ, so verifying these preimages implies all Borel preimages are in 𝒜
CIt is sufficient only on probability spaces, not on general measure spaces
DIt is sufficient because Borel sets are all countable unions of intervals of the form (-∞, a)
The Borel σ-algebra on ℝ is generated by open sets, which in turn are generated by intervals (-∞, a). Since σ-algebras are closed under complements and countable unions, checking that every set {x : f(x) < a} belongs to 𝒜 guarantees that all Borel preimages land in 𝒜 — you don't need to verify each Borel set separately. This is the standard practical criterion for measurability.
Question 3 True / False
Every continuous function from ℝ to ℝ (with the Borel σ-algebra on both sides) is measurable.
TTrue
FFalse
Answer: True
Continuity implies that the preimage of every open set is open, hence Borel, hence measurable. Since the Borel σ-algebra is generated by open sets, all Borel preimages are then measurable. Continuity is strictly stronger than measurability: every continuous function is measurable, but the converse fails.
Question 4 True / False
A measurable function is expected to be continuous at almost nearly every point in its domain.
TTrue
FFalse
Answer: False
Measurability is far weaker than continuity. The Dirichlet function — 1 on rationals, 0 on irrationals — is measurable (it is a pointwise limit of measurable simple functions) yet is continuous nowhere. Measurability is about the preimage structure, not the pointwise behavior of f. Many important measurable functions (indicator functions of Borel sets, for instance) are discontinuous everywhere.
Question 5 Short Answer
Why does the definition of a measurable function require that preimages of Borel sets land in the σ-algebra on the domain, rather than requiring f to satisfy some smoothness or continuity condition?
Think about your answer, then reveal below.
Model answer: The Lebesgue integral works by slicing the range into small intervals and measuring the corresponding preimage sets in the domain. To assign a measure to {x : f(x) ∈ [a, b]}, that set must belong to the σ-algebra 𝒜. The preimage condition is exactly and minimally what is needed: it ensures that every natural question about 'where does f take values in this set?' returns an observable, measurable answer. Smoothness or continuity would be either too restrictive (excluding many integrable functions) or insufficient (they do not guarantee the preimage structure the integral requires).
The definition is not arbitrary formalism. The Lebesgue integral slices the domain via the range, opposite to Riemann's approach. This requires every range-slice to have a measurable domain-preimage. Measurability is precisely the weakest condition that guarantees this, which is why it defines the broadest useful class of integrable functions.