Ex4-15

Question: Let \{f_k\} be a sequence of measurable functions defined on a measurable E with |E|<+\infty. If |f_k(\mathbf{x})|\leq M_\mathbf{x}<+\infty for all k for each \mathbf{x}\in E, show that given \epsilon>0, there is a closed F\subset E and a finite M such that |E-F|<\epsilon and |f_k(\mathbf{x})|\leq M for all k and all \mathbf{x}\in F.
Answer:
Let \epsilon>0. For each n\in\mathbb{N}, define E_n:=\{x\in E: |f_k(x)|\leq n \;\text{for all}\; k\}=\displaystyle\cap_{k=1}^\infty\{x\in E:|f_k(x)|\leq n\} .
Note that each E_n is measurable since each f_k is measurable.
Since each M_x<\infty, we have that E_n\nearrow E. By the Monotone Convergence Theorem for measure, \lim_{n\rightarrow\infty}|E_n|=|E|<\infty. Thus, there exists N such that |E|-|E_N|=|E\backslash E_N|<\epsilon/2.
Let F be a closed set contained in E_N such that |E_N\backslash F|<\epsilon/2.
Then |E\backslash F|=|E\backslash E_N|+|E_N\backslash F|<\epsilon and |f_K(x)|\leq Nfor all k and all x\in F.