(New page: Recall if <math>f\in L^1_{loc}, </math> the result for #3 a follows from Lebesgue differentiation theorem. Next if <math>f\notin L^1_{loc}</math> consider the following: WLOG <math>f\g...) |
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Next if <math>f\notin L^1_{loc}</math> consider the following: | Next if <math>f\notin L^1_{loc}</math> consider the following: | ||
| − | WLOG <math>f\geq 0 </math> by replacing <math>f </math> with <math>|f|.</math> | + | WLOG <math>f\geq 0 </math> by replacing <math> f </math> with <math> |f|.</math> |
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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>\int_kf=\infty</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. | ||
Revision as of 14:52, 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.
