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| − | a/<math>\mu({|f|>0})>0</math> | + | a/<math>\mu({|f|>0})>0</math>, so we have |
| + | <math>(\int_{X}|f|^{n})^{1/n} \leq (\mu(X)||f||_{\infty})^{1/n}</math> | ||
Revision as of 13:30, 11 July 2008
a/$ \mu({|f|>0})>0 $, so we have $ (\int_{X}|f|^{n})^{1/n} \leq (\mu(X)||f||_{\infty})^{1/n} $
