- Assume \(f , f _ { j } \subset L ^ { 2 } ( [ 0 , 1 ] )\) for \(j = 1 , 2 , . . .\) . and \(| | f _ { j } - f | | _ { L ^ { 2 } } \to 0\) . Prove there exists a subsequence \(\{ f _ { j ^ { \prime } } \} \subset \{ f _ { j } \}\) such that \(f _ { j ^ { \prime } } \to f { \mathrm { ~ a . e ~ } }\)
Solution
1. Choose a subsequence with summable squared errors.
Since \[ \|f_j-f\|_2\longrightarrow0, \] we may choose indices \[ j_1<j_2<\cdots \] such that \[ \|f_{j_k}-f\|_2^2<2^{-k} \] for every \(k\ge1\).
2. Apply Tonelli to the sum of the pointwise squared errors.
The summands are nonnegative, so Tonelli’s theorem gives \[ \begin{aligned} \int_0^1\sum_{k=1}^\infty|f_{j_k}(x)-f(x)|^2\,dx &=\sum_{k=1}^\infty\|f_{j_k}-f\|_2^2\\ &<\sum_{k=1}^\infty2^{-k}<\infty. \end{aligned} \] Therefore \[ \sum_{k=1}^\infty|f_{j_k}(x)-f(x)|^2<\infty \] for almost every \(x\in[0,1]\).
3. Conclude pointwise convergence.
Whenever a series of nonnegative terms converges, its individual terms tend to zero. Hence for almost every \(x\), \[ |f_{j_k}(x)-f(x)|^2\longrightarrow0. \] Thus \[ \boxed{f_{j_k}(x)\to f(x)\quad\text{for a.e. }x\in[0,1].} \]
4. Q.E.D.
Step 1 chooses the subsequence, and steps <1>2–<1>3 prove that it converges to \(f\) almost everywhere.