Interchanging limits and Riemann integrals

Try

On [0,2], suppose two integrable functions differ by at most 0.01 everywhere. Their integrals differ by at most 0.02. The interval length converts a height-error bound into an area-error bound. This simple estimate explains why uniform convergence is strong enough to pass a limit through an integral on a fixed compact interval. It also predicts a possible problem when the integration domain has infinite length.

Predict

Predict whether pointwise convergence alone controls area. A spike can become narrower while growing taller, eventually missing every fixed point but retaining the same area. Define f_n=n on (0,1/n) and zero elsewhere in [0,1]. At zero it is explicitly zero. Every fixed positive point is eventually outside the spike, so the pointwise limit is zero, but each integral is one.

Experiment

Inspect the spike mechanism.

n	Height	Width	Integral
2	2	1/2	1
10	10	1/10	1
Any n	n	1/n	1

The table shows exact rectangle areas, not numerical quadrature. A finite grid could miss a narrow spike and report an incorrect integral. Explain why nonnegativity does not rescue this particular pointwise interchange; the missing requirement is a suitable uniform or other domination condition, not the sign alone.

Understand

First prove that a uniform limit f of Riemann-integrable f_n on [a,b] is itself integrable. Uniform convergence makes f bounded: compare it with one bounded f_N within error one. For a uniform error bound η, each partition satisfies U(f,P)≤U(f_N,P)+η(b−a) and L(f,P)≥L(f_N,P)−η(b−a). Hence its sum gap is at most the gap for f_N plus 2η(b−a). For a<b, choose η so the extra term is below ε/2, then choose an integrability partition for f_N with gap below ε/2. Darboux’s criterion gives integrability of f.

Now both integrals exist, and |∫f_n−∫f|≤∫|f_n−f|≤(b−a)||f_n−f||∞. The final bound tends to zero. The proof separates existence of the limit integral from convergence of the integral values; using the latter formula before establishing integrability would omit a necessary step.

Apply

Repair “uniform convergence always lets an improper integral pass through a limit.” On [0,∞), let g_n=1/n on [0,n] and zero elsewhere. Its supremum norm tends to zero, yet its improper integral remains one. The fixed compact-interval estimate has no finite common interval length to use. Tail control must be added separately.

For any fixed R, these same functions do satisfy ∫₀ᴿg_n→0. Convergence on every bounded interval therefore does not control mass moving farther and farther outward. The two failures—mass concentrating into a narrow spike and mass escaping to infinity—are different mechanisms and should be described separately.

Continue

Optional depth points toward stronger integration theorems, such as dominated convergence in measure theory. Those theorems permit some pointwise limits under additional hypotheses, but naming them does not justify an application in a purely Riemann setting without stating their framework and assumptions.

Save one uniform error estimate with the interval length present and one counterexample with exact area calculations. Identify whether the failure comes from lack of uniformity, an infinite domain, or an integrability issue. Next, differentiation will require a substantially stronger convergence condition than integration.