Let \((f_n)\) be Lebesgue-measurable functions on \([0,1]\) satisfying \[ \int_0^1 |f_n(x)|^2\,dx\le \frac1{n^2}. \] Prove that \[ f_n(x)\longrightarrow0 \qquad\text{for almost every }x\in[0,1]. \]
Solution
Fix \(m\in\mathbb N\) and define \[ E_{n,m}=\{x\in[0,1]: |f_n(x)|>1/m\}. \] By Chebyshev’s inequality, \[ \frac1{m^2}m(E_{n,m}) \le \int_{E_{n,m}} |f_n(x)|^2\,dx \le \int_0^1 |f_n(x)|^2\,dx \le \frac1{n^2}, \] so \[ m(E_{n,m})\le \frac{m^2}{n^2}. \] Hence \[ \sum_{n=1}^\infty m(E_{n,m})<\infty. \] By the first Borel–Cantelli lemma, for almost every \(x\) there exists \(N_m(x)\) such that \[ |f_n(x)|\le \frac1m \qquad(n\ge N_m(x)). \] For each \(m\), let \(N_m\) be the null exceptional set where this conclusion fails, and put \[ N=\bigcup_{m=1}^\infty N_m. \] Then \(m(N)=0\). If \(x\notin N\) and \(\varepsilon>0\), choose \(m\) with \(1/m<\varepsilon\). For all \(n\ge N_m(x)\), \[ |f_n(x)|\le\frac1m<\varepsilon. \] Therefore \(f_n(x)\to0\) for every \(x\in[0,1]\setminus N\), proving the assertion.