Let \(\ts{ f_k }_{k=1}^{\infty } \subseteq L^2([0, 1])\) be a sequence which converges in \(L^1\) to a function \(f\).
Prove that \(f\in L^1([0, 1])\).
Give an example illustrating that \(f_k\) may not converge to \(f\) almost everywhere.
Prove that \(\ts{f_k}\) must contain a subsequence that converges to \(f\) almost everywhere.
Solution
1. \(f \in L^1([0,1])\).
Choose \(k\) with \(\|f - f_k\|_1 < 1\). Since \(m([0,1]) = 1\), \(\|f_k\|_1 \le \|f_k\|_2\) by the Cauchy–Schwarz inequality, so \(\|f\|_1 \le \|f - f_k\|_1 + \|f_k\|_1 < \infty\).
2. Let \(E_k\) enumerate the dyadic intervals \([0,\frac12], [\frac12,1], [0,\frac14], [\frac14,\frac12], \ldots\), level by level, and \(f_k = \chi_{E_k}\). Then \(f_k \in L^2([0,1])\) and \(f_k \to 0\) in \(L^1\), but \(f_k(x)\) converges for no \(x \in [0,1]\).
\(|f_k| \le 1\), so \(f_k \in L^2\), and \(\|f_k\|_1 = m(E_k) \to 0\). Each \(x \in [0,1]\) lies in at least one dyadic interval of each level and misses at least one, so \(f_k(x) = 1\) and \(f_k(x) = 0\) for infinitely many \(k\) each.
3. Some subsequence \(f_{k_j} \to f\) a.e.
3.1. There are \(k_1 < k_2 < \cdots\) with \(\|f_{k_j} - f\|_1 < 2^{-j}\).
\(\|f_k - f\|_1 \to 0\).
3.2. For each \(\eps > 0\), \(m\theset{x : |f_{k_j}(x) - f(x)| > \eps \text{ for infinitely many } j} = 0\).
By Markov’s inequality and step 1, \(\sum_j m\theset{|f_{k_j} - f| > \eps} \le \frac1\eps\sum_j 2^{-j} < \infty\), and the Borel–Cantelli lemma applies.
3.3. Q.E.D.
The union over \(\eps = 1/n\), \(n \geq 1\), of the null sets in step 2 is null, and off it \(f_{k_j}(x) \to f(x)\).
4. Q.E.D.
Steps 1, 2 and 3 are parts (a), (b) and (c).