A function φ on [0,1] takes only the values 0, 1, 2, and 3, with each level set {x : φ(x) = k} being a measurable set. What type of function is φ?
AA step function — simple functions require level sets that are intervals, not arbitrary measurable sets
BA simple function — it is a finite linear combination of indicator functions of measurable sets
CA general L¹ function — any bounded function with measurable level sets is integrable but not necessarily simple
DA measurable function but not a simple function — simple functions must take only the value 0 or 1
A simple function is exactly a measurable function that takes finitely many values, each on a measurable set — equivalently, a finite linear combination Σ aᵢ𝟙_{Aᵢ} where each Aᵢ is measurable. The level sets do not need to be intervals; they can be arbitrary measurable sets (Cantor sets, fat Cantor sets, etc.). Options A and D incorrectly restrict simple functions to step functions or indicator functions — those are special cases. Option C correctly identifies integrability but doesn't name the specific function class.
Question 2 Multiple Choice
A student claims that because the increasing sequence φₙ converges to f(x) at every point, for large enough n the approximation becomes uniform — meaning |f(x) − φₙ(x)| < ε for ALL x simultaneously. Why is this claim wrong?
AThe claim is actually correct: pointwise convergence of a monotone increasing sequence implies uniform convergence
BThe claim is wrong because φₙ only converges at points where f is continuous
CThe claim is wrong: pointwise convergence means each fixed x eventually satisfies the bound, but the n that works may differ across points — for an unbounded function, no single n works everywhere simultaneously
DThe claim is wrong because the standard approximating sequence is not monotone increasing
Pointwise convergence guarantees that at each fixed x, you can find N(x) such that n > N(x) implies |f(x) − φₙ(x)| < ε. Uniform convergence requires a single N that works for ALL x simultaneously. For a function growing to infinity (like f(x) = 1/x near 0), the approximation lags arbitrarily far near x = 0 even for large n — those points require ever-larger n. The Common Misconceptions field warns exactly about this: approximation is pointwise, not uniform. Option A is the classic confusion between two distinct modes of convergence.
Question 3 True / False
The standard construction of simple function approximations to a non-negative measurable function f converges uniformly — for large enough n, the supremum of |f(x) − φₙ(x)| over most x approaches zero.
TTrue
FFalse
Answer: False
The approximation is pointwise, not uniform. For each fixed point x, φₙ(x) → f(x) as n → ∞. But for unbounded functions, the supremum sup_x |f(x) − φₙ(x)| may never approach zero — at points where f is very large, the staircase approximation at level n still truncates at n, so the error near a singularity remains large regardless of n. Uniform convergence would require bounding the error simultaneously across all x, which fails for unbounded f.
Question 4 True / False
The standard increasing sequence of simple functions φₙ approximating a non-negative measurable function f satisfies φₙ(x) ≤ φₙ₊₁(x) ≤ f(x) for all x and all n.
TTrue
FFalse
Answer: True
The standard construction assigns each point x the floor of f(x) at the current resolution: at resolution n, assign height k/n where k/n ≤ f(x) < (k+1)/n. Refining to resolution n+1 adds finer height levels, so φₙ₊₁ includes all information from φₙ plus additional detail — the sequence can only increase pointwise. Staying at or below f follows from always assigning the floor, not the ceiling. This monotone structure is what allows the Monotone Convergence Theorem to justify ∫f dμ = lim ∫φₙ dμ.
Question 5 Short Answer
Explain why simple functions serve as the foundation for the Lebesgue integral: what makes them 'simple enough' to integrate by inspection, and what theorem guarantees they are 'rich enough' to approximate any non-negative measurable function?
Think about your answer, then reveal below.
Model answer: Simple functions are finite linear combinations of indicator functions of measurable sets: φ = Σ aᵢ𝟙_{Aᵢ}. Their integral is trivially defined as Σ aᵢμ(Aᵢ) — a weighted sum of finitely many set measures, requiring no limiting arguments. The finiteness is essential: it keeps the integral a straightforward finite sum. The 'richness' side is guaranteed by the approximation theorem: every non-negative measurable function is the pointwise limit of an increasing sequence of simple functions, constructed by discretizing the height axis into strips of width 1/n. The Lebesgue integral for general f is then defined as the limit of these simple-function integrals.
The deep strategy: solve the integration problem on the simplest possible objects (simple functions), then extend by monotone limits — and the theorem guarantees this bridge exists for every non-negative measurable function. This is the Lebesgue approach throughout: rather than integrating arbitrary functions directly, build them as limits of tractable ones.