Let \(f_n\) be a sequence of measurable real-valued functions on \([0,1]\) such that \[ \sum_{n=1}^\infty \int_0^1 |f_n(x)|\,dx\le1. \] Prove that \(f_n\to0\) almost everywhere.
Solution
1. The series \(\sum_n |f_n|\) is finite almost everywhere.
For \(N\ge1\), set \[ S_N(x)=\sum_{n=1}^N |f_n(x)|. \] Then \(S_N\) is measurable and \(S_N\uparrow S:=\sum_{n=1}^\infty |f_n|\). By the monotone convergence theorem, \[ \int_0^1 S(x)\,dx =\lim_{N\to\infty}\int_0^1 S_N(x)\,dx =\sum_{n=1}^\infty\int_0^1 |f_n(x)|\,dx \le1. \] Hence \(S(x)<\infty\) for almost every \(x\in[0,1]\); otherwise its integral would be infinite.
2. Absolute summability implies pointwise convergence to zero.
For every \(x\) for which \(S(x)<\infty\), the numerical series \[ \sum_{n=1}^\infty |f_n(x)| \] converges. Therefore its terms tend to zero: \[ |f_n(x)|\longrightarrow0. \] Thus \(f_n(x)\to0\) for almost every \(x\in[0,1]\).