- Is it true that the converse to the DCT holds? I.e. if \(\int f_n \to \int f\), is there a \(g\in L^p\) such that \(f_n < g\) a.e. for every \(n\)?
Solution
The answer is no. Let \(f \equiv 0\) and \(f_n = n\chi_{[1/(n+1), 1/n)}\) on \(\RR\) for \(n \ge 1\).
1. \(f_n \geq 0\) and \(f_n \to 0\) pointwise.
For \(x \leq 0\) or \(x \geq 1\), \(f_n(x) = 0\) for all \(n\). For \(0 < x < 1\), \(f_n(x) = 0\) once \(1/n < x\).
2. \(\int f_n = n\left(\frac{1}{n} - \frac{1}{n+1}\right) = \frac{1}{n+1} \to 0 = \int f\).
This is the integral of a constant over an interval of length \(\frac1n - \frac1{n+1}\).
3. \(\int \sup_n f_n = \infty\).
The intervals \([1/(n+1), 1/n)\) are pairwise disjoint and \(f_n\) is supported on its own interval, so \(\sup_n f_n = n\) on \([1/(n+1), 1/n)\); hence \(\int \sup_n f_n = \sum_{n=1}^\infty n \cdot \frac{1}{n(n+1)} = \sum_{n=1}^\infty \frac{1}{n+1} = \infty\).
4. No \(g \in L^p\) with \(1 \leq p < \infty\) satisfies \(f_n \leq g\) a.e. for every \(n\).
Such a \(g\) satisfies \(g \ge \sup_n f_n \geq 0\) a.e., so \(\int g^p \ge \int (\sup_n f_n)^p = \sum_n n^p \cdot \frac{1}{n(n+1)} \ge \frac{1}{2}\sum_n n^{p-2} = \infty\).
5. Q.E.D.
By steps 1 and 2, \(f_n \to f\) pointwise and \(\int f_n \to \int f\), while step 4 shows that no \(L^p\) function dominates the \(f_n\).