Let \((X, \mathcal{M}, \mu)\) be a measure space and \(0 < p < q< \infty\). Prove that if \(L^q(X) \subseteq L^p(X)\), then \(X\) does not contain sets of arbitrarily large finite measure.
Solution
We prove the contrapositive. Assume \(X\) contains measurable sets of arbitrarily large finite measure.
1. There are pairwise disjoint measurable \(E_n\), \(n \ge 1\), with \(2^n \le \mu(E_n) < \infty\).
Inductively choose a measurable \(B_n\) with \(2^n + \sum_{k<n} \mu(E_k) \le \mu(B_n) < \infty\) and put \(E_n = B_n \setminus \bigcup_{k<n} E_k\). Then \(\mu(E_n) \ge \mu(B_n) - \sum_{k<n}\mu(E_k) \ge 2^n\), and \(\mu(E_n) \le \mu(B_n) < \infty\).
2. \(f = \sum_n \mu(E_n)^{-1/p}\chi_{E_n}\) lies in \(L^q \setminus L^p\).
The \(E_n\) are disjoint, so \(\int f^q = \sum_n \mu(E_n)^{1 - q/p} \le \sum_n 2^{n(1-q/p)} < \infty\) because \(q/p > 1\), while \(\int f^p = \sum_n 1 = \infty\).
3. Q.E.D.
Step 2 shows \(L^q \not\subseteq L^p\).