5-23

Question:
Prove the following fact, sometimes referred to as the Sequential (or Generalized) Version of the Lebesgue Dominated Convergence Theorem. Let \{f_k\} and \{\phi_k\} be sequences of measurable functions on E satisfying f_k \to fa.e. in E, \phi_k \to \phi a.e. in E, and |f_k|\leq \phi_k a.e. in E. If \phi \in L(E) and \int_E\phi_k \to\int_E \phi , then \int_E |f_k-f| \to 0.
Solution:
# We first note that \phi_k \in L(E) for large enough k and so f_k \in L(E) for large enough k. This is because there exists K such that for all k\geq K, 0 \leq \int_E\phi_k < \left(\int_E \phi\right)+1 < \infty.It follows that all the integrals involved are finite and the linearity of integral holds.
# Since |f_k|\leq \phi_k, it follows that |f|=\liminf |f_k|\leq \liminf \phi_k =\phi. So f \in L(E).
# By the triangle inequality, |f_k-f|\leq \phi_k+\phi. Applying Fatou's Lemma to the nonnegative sequence of functions \{\phi_k+\phi-|f_k-f|\}, we obtain\begin{align}
\int_E \liminf (\phi_k+\phi-|f_k-f|) &\leq \liminf \int_E \phi_k+\phi-|f_k-f|\\
\int_E 2\phi &\leq 2\int_E \phi + \liminf (-\int_E|f_k-f|)\\
\limsup \int_E |f_k-f| &\leq 0.
\end{align}Since \int_E |f_k-f|\geq 0, \liminf \int_E |f_k-f|\geq 0. Hence, \lim_{k \to \infty} |f_k-f|=0.