| Line 6: | Line 6: | ||
Let <math>x\in \mathbb{R}^n</math>. | Let <math>x\in \mathbb{R}^n</math>. | ||
| − | '''Case 1''', <math>\exists K\subset \mathbb{R}^n, K </math> compact, and<math>\ | + | '''Case 1''', <math>\exists K\subset \mathbb{R}^n, K </math> compact, and<math>\int_Kf=\infty</math>. |
| − | Choose a cube <math>Q\supseteq K</math> with <math>|Q|<\infty </math> which is possible since <math>K</math> compact implies <math>K </math> bounded. | + | Choose a cube <math>Q\supseteq K</math> with <math>|Q|<\infty </math> which is possible since <math>K</math> compact implies <math>K </math> bounded. WLOG <math>x</math> is the center of <math>Q.</math> |
| + | Then <math>f^*(x)\geq\dfrac{\int_Qf}{|Q|} \geq \dfrac{\int_Kf}{|Q|}=\infty</math>, so our result holds on <math> \{x|\exists K s.t. \intKf=\infty \} </math>. | ||
Revision as of 13:57, 9 July 2008
Recall if $ f\in L^1_{loc}, $ the result for #3 a follows from Lebesgue differentiation theorem.
Next if $ f\notin L^1_{loc} $ consider the following: WLOG $ f\geq 0 $ by replacing $ f $ with $ |f|. $
Let $ x\in \mathbb{R}^n $.
Case 1, $ \exists K\subset \mathbb{R}^n, K $ compact, and$ \int_Kf=\infty $. Choose a cube $ Q\supseteq K $ with $ |Q|<\infty $ which is possible since $ K $ compact implies $ K $ bounded. WLOG $ x $ is the center of $ Q. $ Then $ f^*(x)\geq\dfrac{\int_Qf}{|Q|} \geq \dfrac{\int_Kf}{|Q|}=\infty $, so our result holds on $ \{x|\exists K s.t. \intKf=\infty \} $.
