- For each natural number n, let \(f _ { n } : [ 0 , 1 ] \to \mathbb { R }\) be a sequence of absolutely integrable functions, and let \(f : [ 0 , 1 ] \to\) R be another absolutely integrable function such that
\[ \int _ { 0 } ^ { 1 } { \big | } f _ { n } ( x ) - f ( x ) { \big | } d x \to 0 , \qquad { \mathrm { a s } } \quad n \to \infty . \]
Show that there exists a subsequence \(f _ { n _ { j } }\) of \(f _ { n }\) which converges to \(f\) pointwise almost everywhere.
Give a counterexample to show that the assertion fails if ”pointwise almost everywhere” is replaced by ”uniformly”.
Solution
1. Choose a subsequence with summable \(L^1\) errors.
Since \[ \|f_n-f\|_1\longrightarrow0, \] we may choose indices \[ n_1<n_2<\cdots \] such that \[ \|f_{n_j}-f\|_1<2^{-j} \] for every \(j\).
2. Prove almost-everywhere convergence of that subsequence.
By Tonelli’s theorem, \[ \begin{aligned} \int_0^1\sum_{j=1}^\infty |f_{n_j}(x)-f(x)|\,dx &=\sum_{j=1}^\infty\|f_{n_j}-f\|_1\\ &<\sum_{j=1}^\infty2^{-j}<\infty. \end{aligned} \] Hence \[ \sum_{j=1}^\infty|f_{n_j}(x)-f(x)|<\infty \] for almost every \(x\). Therefore \[ |f_{n_j}(x)-f(x)|\longrightarrow0 \] for almost every \(x\), proving part (a).
3. Give a counterexample to uniform convergence.
Let \[ f_n=\mathbf1_{(0,1/n)}, \qquad f=0. \] Then \[ \|f_n-f\|_1=\frac1n\longrightarrow0. \] However, \[ \|f_n-f\|_\infty=1 \] for every \(n\). The same is true for every subsequence, so no subsequence can converge uniformly to \(f\).
Thus the almost-everywhere conclusion cannot be replaced by uniform convergence.