Prove that if $f_n\rightarrow f$ in $L^{\infty }(\lambda )$. Then $f_n\rightarrow f$ UNIFORMLY on a measurable set whose complement has $\mu-$measure zero.
Prove that if $f_n\rightarrow f$ in $L^{\infty }(\lambda )$. Then $f_n\rightarrow f$
0
$\begingroup$
measure-theory
1 Answers
2
Let $\Omega$ be the domain. Then $f_{n}\rightarrow f$ in $L^{\infty}$ means that $\|f_{n}-f\|_{\infty}=\mathrm{esssup}_{x\in\Omega}|f_{n}(x)-f(x)|\rightarrow0$ as $n\rightarrow\infty.$ For each $n,$ $E_{n}=\{x\in\Omega:|f_{n}(x)-f(x)|>\|f_{n}-f\|_{\infty}\}$ has $\lambda$-measure 0 by definition of ess sup. Let $A=\Omega\setminus\left(\bigcup_{n=1}^{\infty}E_{n}\right).$ Then $\lambda(A^{c})\leq\sum_{n=1}^{\infty}\lambda(E_{n})=0,$ and $\sup_{x\in A}|f_{n}(x)-f(x)|\leq\|f_{n}-f\|_{\infty}\rightarrow 0,$ so $f_{n}\rightarrow f$ uniformly on $A.$