| Line 6: | Line 6: | ||
Since <math> f \notin L^1_{loc}, \exists x\in \mathbb{R}^n and \exists K\subset \mathbb{R}^n, K </math> compact,s.t. <math>\int_Kf=\infty</math>. | Since <math> f \notin L^1_{loc}, \exists x\in \mathbb{R}^n and \exists K\subset \mathbb{R}^n, K </math> compact,s.t. <math>\int_Kf=\infty</math>. | ||
Choose any <math>y\in\mathbb{R}^n </math>. Then choose a cube <math>Q\supseteq K</math> centered at <math>y </math> which is possible since <math>K</math> compact implies <math>K </math> bounded. | Choose any <math>y\in\mathbb{R}^n </math>. Then choose a cube <math>Q\supseteq K</math> centered at <math>y </math> which is possible since <math>K</math> compact implies <math>K </math> bounded. | ||
| − | Then <math> f^*(y)\geq \dfrac{\int_Qf}{|Q|} \geq \dfrac{\int_Kf}{|Q|}=\infty </math>, so we have the result, namely <math> \{ f* \geq f \} =\mathbb{R}^n </math> | + | Then <math> f^*(y)\geq \dfrac{\int_Qf}{|Q|} \geq \dfrac{\int_Kf}{|Q|}=\infty </math>, so we have the result, namely <math> \{ f^* \geq f \} =\mathbb{R}^n </math> |
QED | QED | ||
Latest revision as of 14:09, 9 July 2008
Recall if $ f\in L^1_{loc}, $ the result for #3a follows from Lebesgue differentiation theorem.
Next if $ f\notin L^1_{loc} $ consider the following proof: WLOG $ f\geq 0 $ by replacing $ f $ with $ |f|. $
Since $ f \notin L^1_{loc}, \exists x\in \mathbb{R}^n and \exists K\subset \mathbb{R}^n, K $ compact,s.t. $ \int_Kf=\infty $. Choose any $ y\in\mathbb{R}^n $. Then choose a cube $ Q\supseteq K $ centered at $ y $ which is possible since $ K $ compact implies $ K $ bounded. Then $ f^*(y)\geq \dfrac{\int_Qf}{|Q|} \geq \dfrac{\int_Kf}{|Q|}=\infty $, so we have the result, namely $ \{ f^* \geq f \} =\mathbb{R}^n $
QED
