Suppose \(f_n:[0,1]\to\mathbb{R}\) are continuous functions converging uniformly to \(f:[0,1]\to\mathbb{R}\). Either prove that \(\displaystyle\lim_{n\to\infty}\int_{1/n}^1 f_n(x)\,dx=\int_0^1 f(x)\,dx\) or give a counterexample.
Solution
1. Claim: \(\lim_{n\to\infty}\int_{1/n}^1 f_n(x)\,dx = \int_0^1 f(x)\,dx\).
split the difference into the convergence part (uniform) and the endpoint part (small interval).
2. \(\int_{1/n}^1 f_n = \int_{1/n}^1 f + \int_{1/n}^1 (f_n - f)\), and \(\left|\int_{1/n}^1 (f_n - f)\right| \le \|f_n - f\|_\infty \cdot 1 \to 0\).
uniform convergence \(f_n \to f\) means \(\|f_n - f\|_\infty \to 0\); the interval has length \(\le 1\).
3. \(\int_{1/n}^1 f \to \int_0^1 f\).
\(f\) is continuous on \([0,1]\) (uniform limit of continuous functions), hence bounded, \(|f| \le M\); then \(\left|\int_0^{1/n} f\right| \le M/n \to 0\), so \(\int_{1/n}^1 f = \int_0^1 f - \int_0^{1/n} f \to \int_0^1 f\).
4. Q.E.D.
2 and 3 combine: \(\int_{1/n}^1 f_n = \int_{1/n}^1 f + o(1) \to \int_0^1 f\).