If \(f:[0,1]\to\mathbb{R}\) is continuous, prove that \[\displaystyle\lim_{n\to\infty}\int_0^1 f(x^n)\,dx=f(0).\]
Solution
1. \(f(x^n) \to f(0)\) pointwise for every \(x \in [0,1)\).
\(x^n \to 0\) for \(0 \le x < 1\); \(f\) is continuous, so \(f(x^n) \to f(0)\).
2. \(f(x^n) \to f(0)\) for almost every \(x \in [0,1]\).
1 covers all \(x\) except \(x = 1\), a null set.
3. The family is dominated: \(|f(x^n)| \le \|f\|_\infty\) for all \(n\) and all \(x\).
\(f\) is continuous on the compact interval \([0,1]\), hence bounded; \(x^n \in [0,1]\).
4. Q.E.D.
dominated convergence (<1>2, 3) gives \(\lim_n\int_0^1 f(x^n)\,dx = \int_0^1 f(0)\,dx = f(0)\).