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Spring 2020 - Problem 3

measure theory

Prove that L∞(Rn)∩L3(Rn)L^\infty\p{\R^n} \cap L^3\p{\R^n} is a Borel subset of L3(Rn)L^3\p{\R^n}. Let FN={f∈L3(Rn) | ∥f∥L∞≤N}F_N = \set{f \in L^3\p{\R^n} \st \norm{f}_{L^\infty} \leq N}. Then

L∞(Rn)∩L3(Rn)=⋃N=1∞FN,L^\infty\p{\R^n} \cap L^3\p{\R^n} = \bigcup_{N=1}^\infty F_N,

so it suffices to show that each FNF_N is closed. Suppose {fk}k⊆FN\set{f_k}_k \subseteq F_N converges to ff in L3(Rn)L^3\p{\R^n}. Then passing to a subsequence, we may assume that fk→ff_k \to f pointwise almost everywhere. Observe then that

⋃k=1∞{x∈Rn | ∣f(x)∣≤N}∪{x∈Rn | fk(x)→f(x)}\bigcup_{k=1}^\infty \set{x \in \R^n \st \abs{f\p{x}} \leq N} \cup \set{x \in \R^n \st f_k\p{x} \to f\p{x}}

is a full measure set as a countable union of full measure sets. In other words, for almost every x∈Rnx \in \R^n, we have

∣f(x)∣≤sup⁡k∣fk(x)∣≤N,\abs{f\p{x}} \leq \sup_k \abs{f_k\p{x}} \leq N,

so f∈FNf \in F_N, which was what we wanted to show.
Solution.