5-26

Prove the following variant of Lebesgue's dominated convergence theorem: if \{f_k\} satisfies f_k {\overset{m}{\longrightarrow}} f on E and |f_k| \le \phi \in L(E), then f \in L(E) and \displaystyle \int_E f_k \to \int_E f. (Show that every subsequence of \{f_k\} has a subsequence \{f_{k_j}\} such that \displaystyle\int_E f_{k_j} \to \int_E f.)
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We prove by contradiction, that is, suppose \displaystyle \int_E f_k \not\to \int_E f. Then there exists \varepsilon > 0 and a subsequece \{f_{k_j}\} of \{f_k\} such that
\left| \int_E f_{k_j} - \int_E f \right| \ge \varepsilon
for all j \in \Z^+.
Since f_k {\overset{m}{\longrightarrow}} f on E, then so is f_{k_j} {\overset{m}{\longrightarrow}} f on E. By Theorem 4.22, there is a subsequence \{f_{k_{j_l}}\} of \{f_{k_j}\} such that f_{k_{j_l}} \to f a.e. in E. Thus by applying Theorem 5.36 to \{f_{k_{j_l}}\}, we have \displaystyle\int_E f_{k_{j_l}} \to \int_E f, a contradiction.