Consider the sequence fₙ = 1_{[n, n+1]} (the indicator function of the interval [n, n+1] on ℝ). This sequence converges pointwise to 0 everywhere, yet ∫fₙ dμ = 1 for all n. Why does the Dominated Convergence Theorem fail to apply here?
AThe sequence does not converge pointwise — it only converges in measure
BThere is no integrable dominating function on ℝ for this sequence; the natural candidate g = 1 has infinite Lebesgue integral
CThe functions fₙ are not measurable because they change domain with each n
DDCT requires uniform convergence, and this sequence only converges pointwise
The DCT requires both pointwise a.e. convergence and an integrable dominating function g with |fₙ| ≤ g a.e. Both conditions are load-bearing. For this sequence, pointwise convergence holds (each fₙ(x) = 0 eventually for any fixed x, since x ∉ [n, n+1] for large n). But the natural dominating function would be g = 1, and ∫ℝ 1 dμ = ∞ — g is not integrable on ℝ. There is no integrable dominator: the mass escapes to infinity rather than being controlled. This is precisely the intuition DCT formalizes — an integrable dominator prevents the sequence's integral from drifting away even while individual values go to zero.
Question 2 Multiple Choice
You want to justify differentiating under the integral sign: d/dt ∫₀¹ f(x,t) dx = ∫₀¹ ∂f/∂t(x,t) dx. Which condition, when satisfied, allows you to apply DCT to justify this exchange?
Af(x,t) is continuous in t for each fixed x ∈ [0,1]
BThe partial derivative ∂f/∂t exists everywhere and is bounded on [0,1] × [a,b]
CThere exists an integrable function g(x) on [0,1] such that |∂f/∂t(x,t)| ≤ g(x) for all t in the relevant range
Df(x,t) is a uniformly convergent sequence of simple functions in t
Differentiating under the integral sign is justified by applying DCT to the difference quotients (f(x, t+h) - f(x, t))/h as h → 0. For DCT to apply, these difference quotients must be dominated by an integrable function. By the mean value theorem, |(f(x, t+h) - f(x,t))/h| ≤ sup |∂f/∂t(x,τ)|. If there is an integrable g(x) with |∂f/∂t(x,t)| ≤ g(x) for all relevant t, then DCT applies and the limit (the derivative) can pass through the integral. Mere continuity (option A) or pointwise boundedness without integrability (option B) is insufficient — the dominator must be integrable.
Question 3 True / False
If fₙ is a sequence of measurable functions on a set E of finite measure, and |fₙ(x)| ≤ M for all x and all n (a bounded sequence), then g(x) = M · 1_E(x) is an integrable dominating function, making the DCT applicable whenever fₙ converges pointwise a.e.
TTrue
FFalse
Answer: True
On a finite-measure set with a uniformly bounded sequence, finding an integrable dominator is automatic: g = M · 1_E satisfies |fₙ| ≤ g pointwise and ∫g dμ = M · μ(E) < ∞ because μ(E) is finite. This is why DCT is 'almost automatic' for bounded sequences on finite-measure domains — the two hardest parts (finding g and verifying its integrability) reduce to a trivial observation. The real analytical challenge arises on infinite-measure domains or for unbounded sequences, where finding a creative integrable dominator is the non-trivial work.
Question 4 True / False
Pointwise convergence almost everywhere of a sequence of integrable functions is sufficient to justify exchanging the limit with the Lebesgue integral: lim ∫fₙ = ∫ lim fₙ.
TTrue
FFalse
Answer: False
Pointwise convergence alone is not sufficient — this is exactly what the sequence fₙ = n · 1_{(0, 1/n)} demonstrates. Each fₙ integrates to 1 (the integral of n over an interval of length 1/n is 1), yet fₙ(x) → 0 pointwise almost everywhere (for any fixed x > 0, fₙ(x) = 0 eventually). The limit of integrals is 1 but the integral of the limit is 0. The DCT requires the additional condition of an integrable dominating function to prevent this kind of 'mass concentration' — where functions spike near a point with growing height and shrinking support, carrying finite area while converging to zero pointwise.
Question 5 Short Answer
What role does the dominating function g play in the DCT, and why must g be integrable (∫g < ∞) rather than merely finite pointwise (g(x) < ∞ for a.e. x)?
Think about your answer, then reveal below.
Model answer: The dominating function g acts as a uniform leash on the entire sequence: |fₙ(x)| ≤ g(x) for all n means no function in the sequence can put more mass anywhere than g does. Integrability (∫g < ∞) is what prevents mass from escaping to infinity in the limit — it ensures that the total 'budget' of mass the sequence can deploy is finite. If g is merely finite pointwise but has infinite integral, the sequence can redistribute mass without limit even while converging pointwise. The example fₙ = 1_{[n,n+1]} shows this: each function is bounded by 1 everywhere (pointwise finite), but the mass keeps moving to infinity, so ∫fₙ = 1 while ∫(lim fₙ) = 0. The integrability of g is the condition that prevents this escape.
The intuition is that ∫g < ∞ constrains not just the height of the functions but the total weight they can carry. Fatou's lemma (a weaker result requiring no dominator) gives only lim inf ∫fₙ ≥ ∫(lim inf fₙ) — an inequality, not equality. The dominator is what converts the inequality into equality, by blocking the mass from escaping to sets where the limit function is small.