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First, if <math> m(X)= \infty </math>, it's done. Hence let's suppose that <math> m(X)<\infty </math> | First, if <math> m(X)= \infty </math>, it's done. Hence let's suppose that <math> m(X)<\infty </math> | ||
| − | Now, WTS that <math> f \in L^{p} </math>. | + | Now, WTS that <math> f \in L^{p} </math>, which is equivalent to show that <math> |f|^p \in L^{1} </math> |
| + | |||
| + | Let <math> D_n=\{x \in X : |f(x)| \geq n \}</math>. Then <math> \sum_{n=0}^{\infty}m(D_n)=\sum_{n=0}^{\infty}(n+1)m(E_n)</math>. Thus, | ||
| + | |||
| + | <math> \sum_{n=0}^{\infty}m(D_n)=\sum_{n=0}^{\infty}(n+1)m(E_n)=\sum_{n=0}^{\infty}m(E_n)+\sum_{n=0}^{\infty}nm(E_n)=m(X)+\sum_{n=0}^{\infty}nm(E_n)</math>. | ||
Revision as of 23:15, 10 July 2008
Suppose we know the conclusion of problem 8,
Problem 8 Let $ X $ be a finite measure space. If $ f $ is measurable, let
$ E_n = \{x \in X : n-1 \leq |f(x)| < n \} $. Then
$ f \in L^1 $ if and only if $ \sum_{n=1}^{\infty}nm(E_n) < \infty. $
First, if $ m(X)= \infty $, it's done. Hence let's suppose that $ m(X)<\infty $
Now, WTS that $ f \in L^{p} $, which is equivalent to show that $ |f|^p \in L^{1} $
Let $ D_n=\{x \in X : |f(x)| \geq n \} $. Then $ \sum_{n=0}^{\infty}m(D_n)=\sum_{n=0}^{\infty}(n+1)m(E_n) $. Thus,
$ \sum_{n=0}^{\infty}m(D_n)=\sum_{n=0}^{\infty}(n+1)m(E_n)=\sum_{n=0}^{\infty}m(E_n)+\sum_{n=0}^{\infty}nm(E_n)=m(X)+\sum_{n=0}^{\infty}nm(E_n) $.
