5)
Fix $ \alpha > 0 $.
Define $ f_{n_k} $ inductively as follows.
Let $ f_{n_{0}}=f_1 $
Let $ \epsilon_1 = 1 $
Let $ A_1=[0,\alpha+1] $.
Now suppose $ f_{n_k} $ is defined for $ 0<k<K $. Further suppose that the set $ A_K $ is given with $ m(A_K) \leq \alpha +1/K $ and the number $ \epsilon_K $ is given.
Then by Egorov, I can find a compact set $ E_K \subset A_K $ with $ m(E_K) > \alpha + 1/(K+1) $ such that $ f_n \rightarrow $ 0 uniformly on $ E_K $.
Then $ \exists \ N>n_{K-1} s.t. \ n\geq N \Rightarrow f_n< \epsilon_K. $ Define $ f_{n_K}=f_{N} $.
Now since $ m(E_K)> \alpha + 1/(K+1) $, and since $ f_{n_K}>0 $, we can find $ \epsilon_{K+1}>0 $ such that $ m(\{f_{n_K}>\epsilon_{K+1} \} \cap E_K)> \alpha + 1/(K+1) $. Define
$ A_{K+1}=\{f_{n_K}>\epsilon_{K+1} \} \cap E_K $.
So, since $ A_{k+1} $ and $ \epsilon_{K+1} $ are defined and since $ m(A_{k+1}) > \alpha + 1/(K+1) $, we can then define $ f_{n_{K+1}} $, this process defines inductively our subsequence $ f_{n_k} $.
Now let $ E = \cap_{n=1}^\infty E_n $. Then since the $ {E_n} $ is a decreasing sequence of compact sets each with measure $ > \alpha, \ E $ has measure $ \geq \alpha $. Thus, $ \exists \ $ a measurable set $ A \subset E \ m(A)=\alpha $. Then for $ k\geq1 $, and $ x \in A, f_{n_k}(x)>\epsilon_{k+1}>f_{n_{k+1}}(x) $. So this set $ A $, and the subsequence $ \{f_{n_k}\}_{k=1}^\infty $ satisfies the requirements.
--Wardbc 16:37, 8 July 2008 (EDT)
