Give a sequence of functions on [0, 1] that converges pointwise to zero everywhere while the integral of each one equals 1. Name the hypothesis of the dominated convergence theorem that your example breaks, and say what monotone convergence would need instead.

Give a sequence of functions on [0, 1] that converges pointwise to zero everywhere while the integral of each one equals 1. Name the hypothesis of the dominated convergence theorem that your example breaks, and say what monotone convergence would need instead.

Approach: Build a spike whose height grows exactly as fast as its width shrinks, then compute the pointwise supremum over the whole sequence and test whether that envelope is integrable.

Take f_n(x) = n on (0, 1/n) and 0 elsewhere. For any fixed x > 0 there is an N with 1/N < x, so f_n(x) = 0 for all n >= N, giving pointwise convergence to zero, and f_n(0) = 0 for every n. Each integral is n * (1/n) = 1, so the limit of the integrals is 1 while the integral of the limit is 0. The dominated convergence theorem fails on its integrable envelope: the smallest dominating function is sup_n f_n(x), which equals n on (1/(n+1), 1/n] and behaves like 1/x, and the integral of 1/x on (0, 1] diverges. Monotone convergence does not apply either, since it needs an increasing sequence of non-negative functions and this one is not monotone in n. The escaping mass is the general failure mode, and any hypothesis that pins the mass down, an integrable envelope or uniform integrability, restores the interchange.

Follow-up: Construct a sequence converging to zero pointwise whose integrals diverge to infinity, and one where the integrals oscillate without limit?

Key concepts: dominated convergence theorem, pointwise convergence, integrable envelope, monotone convergence.