Consider a sequence of functions fₙ on [0,1] where fₙ = n on [0, 1/n] and fₙ = 0 elsewhere. Each fₙ converges pointwise to f = 0. What is lim(n→∞) ∫₀¹ fₙ dx?
A0, because fₙ → 0 pointwise and the limit of a continuous function is integrable
B1, because each ∫fₙ = 1 regardless of the pointwise limit
C∞, because the spikes grow without bound
DUndefined, because the integral does not exist for each fₙ
Each fₙ has integral exactly 1 (height n times width 1/n). Yet fₙ(x) → 0 for every fixed x in [0,1], because for any x > 0, fₙ(x) = 0 once n > 1/x. So ∫f = ∫0 = 0, but lim ∫fₙ = 1 ≠ 0 = ∫ lim fₙ. This is the canonical counterexample: pointwise convergence allows the integral mass to 'escape' into a shrinking spike, so the interchange fails. Uniform convergence would prevent this by requiring the error to shrink simultaneously everywhere.
Question 2 Multiple Choice
Under which condition is it guaranteed that lim(n→∞) ∫ₐᵇ fₙ dx = ∫ₐᵇ (lim fₙ) dx on a bounded interval [a,b]?
AEach fₙ is continuous and the sequence is monotone increasing
Bfₙ converges pointwise to f and each fₙ is bounded by the same constant M
Cfₙ converges uniformly to f on [a,b] and each fₙ is integrable
DThe sequence is Cauchy in the sup-norm at every rational point of [a,b]
Uniform convergence on [a,b] is the key hypothesis in the classical interchange theorem. It allows a single ε to work for all x simultaneously, so |∫fₙ − ∫f| ≤ ∫|fₙ − f| ≤ ε·(b−a). Option A (monotone) is the hypothesis for the Monotone Convergence Theorem, which applies to Lebesgue integrals in a different setting. Option B (uniform bound) is the hypothesis for the Dominated Convergence Theorem, not the classical Riemann-based result. Option D has a real-analysis flavor but is not a standard sufficient condition.
Question 3 True / False
If (fₙ) converges uniformly to f on [a,b] and each fₙ is Riemann integrable, then lim ∫ₐᵇ fₙ dx = ∫ₐᵇ f dx.
TTrue
FFalse
Answer: True
True. This is the uniform convergence interchange theorem. The proof is direct: |∫fₙ − ∫f| = |∫(fₙ − f)| ≤ ∫|fₙ − f| ≤ sup_x|fₙ(x) − f(x)| · (b−a). Since fₙ → f uniformly, the sup-norm goes to 0, so the difference between the integrals goes to 0. Note that uniform convergence also implies f is integrable (as a uniform limit of integrable functions on a bounded interval is itself integrable).
Question 4 True / False
Pointwise convergence of (fₙ) to f on [a,b] is sufficient to conclude that lim ∫ₐᵇ fₙ dx = ∫ₐᵇ f dx.
TTrue
FFalse
Answer: False
False. The 'spike' counterexample (fₙ = n on [0,1/n], 0 elsewhere) shows pointwise convergence is not enough: fₙ → 0 pointwise everywhere, yet each integral equals 1. With pointwise convergence, the error |fₙ(x) − f(x)| need not shrink simultaneously across all x — it can concentrate in ever-smaller sets that still contribute finite area. Uniform convergence prevents this by requiring the error to be controlled everywhere at once.
Question 5 Short Answer
Explain in your own words why uniform convergence is the 'right' condition for interchanging limit and integral, when pointwise convergence is not.
Think about your answer, then reveal below.
Model answer: With pointwise convergence, the error |fₙ(x) − f(x)| can be small at each individual x but large somewhere on the interval — 'moving' the mass around as n grows. This lets the integral of fₙ differ from the integral of f. Uniform convergence requires the error to be small everywhere simultaneously: once n is large enough, |fₙ(x) − f(x)| < ε for all x at once. Then the integral error is at most ε times the interval length, which can be made arbitrarily small.
The intuition is that integration is a global operation — it sums contributions from the whole interval. Pointwise convergence only gives local control (at each fixed point), which is too weak to control the global sum. Uniform convergence gives the global control needed. In more advanced analysis, the Dominated Convergence Theorem gives a weaker condition (domination by an integrable function) that covers more cases while still maintaining enough global control.