Ex4-16

Question: Prove that f_k\xrightarrow{m}f on E if and only if given \epsilon>0, there exists K such that |\{|f-f_k|>\epsilon\}|<\epsilon if k>K. Give an analogous Cauchy criterion.
Answer:
(\Longrightarrow)Assume f_k\xrightarrow{m}f on E. Let \epsilon>0 be given. By definition of convergence in measure, \lim_{x\rightarrow\infty}|\{x\in E:|f(x)-f_k(x)|>\epsilon\}|=0.
By definition of limit, there exists K such that |\{|f-f_k|>\epsilon\}|<\epsilon if k>K.
(\Longleftarrow)Assume that given any \epsilon>0, there exists K such that |\{|f-f_k|>\epsilon\}|<\epsilon if k>K.
Let \delta>0. Given any \epsilon>0, we wish to show that there exists K such that |\{|f-f_k|>\delta\}|<\epsilon if k<K. This would imply \lim_{k\rightarrow\infty}|\{|f-f_k|>\delta\}|=0, and thus f_k\xrightarrow{m}f.
case 1 : \epsilon\geq\delta.
By assumption, there exists K such that |\{|f-f_k|>\delta\}|<\delta\leq\epsilon if k>K.
case 2 : \epsilon<\delta.
Since \{|f-f_k|>\delta\}\subseteq\{|f-f_k|>\epsilon\}, thus |\{|f-f_k|>\delta\}|\leq|\{|f-f_k|>\epsilon\}|<\epsilon if k>K.
Give an analogous Cauchy criterion: f_k\xrightarrow{m}f on E if and only if given \epsilon>0, there exists K such that |\{|f_k-f_l|>\epsilon\}|<\epsilon if k,l>K.
Proof is analogous to the previous part, using Theorem 4.23 (Cauchy criterion for convergence in measure).