Let \(1\le p<\infty\) and suppose \(f_n\to f\) in \(L^p([0,1])\).
Prove that \(f_n\to f\) in measure.
Prove that some subsequence converges to \(f\) almost everywhere.
Solution
1. Convergence in measure.
Fix \(\varepsilon>0\). By Chebyshev’s inequality, \[ \mu\{|f_n-f|\ge\varepsilon\} \le \frac1{\varepsilon^p}\int_0^1|f_n-f|^p\,dx. \] The right-hand side tends to \(0\) because \(f_n\to f\) in \(L^p\). Hence \(f_n\to f\) in measure.
2. An almost-everywhere convergent subsequence.
Choose a subsequence \((f_{n_k})\) such that \[ \|f_{n_k}-f\|_p^p\le2^{-k}. \] Then \[ \int_0^1\sum_{k=1}^\infty |f_{n_k}(x)-f(x)|^p\,dx =\sum_{k=1}^\infty\|f_{n_k}-f\|_p^p \le1. \] By Tonelli’s theorem, \[ \sum_{k=1}^\infty |f_{n_k}(x)-f(x)|^p<\infty \] for almost every \(x\). Therefore its terms tend to zero, so \[ f_{n_k}(x)\longrightarrow f(x) \] for almost every \(x\in[0,1]\).