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| + | We know that <math>\int_X |f|^r = \int_0^\infty ry^{r-1}\mu\left\{|f|>y\right\} dy</math>. If <math>\mu(X) = \infty</math>, the statement is trivial, so assume it's finite. | ||
| + | We have | ||
| + | |||
| + | <math>\int_0^\infty ry^{r-1}\mu(|f|>y) dy = \int_0^{\mu(X)^{\frac{-1}{p}}} ry^{r-1}\mu\left\{|f|>y\right\} dy | ||
| + | +\int_{\mu(X)^{\frac{-1}{p}}}^\infty ry^{r-1}\mu\left\{|f|>y\right\} dy </math> | ||
| + | |||
| + | Now <math>\mu(|f|>y)\leq\mu(X) \Rightarrow \int_0^{\mu(X)^{\frac{-1}{p}}} ry^{r-1}\mu(|f|>y) dy \leq \mu(X)\int_0^{\mu(X)^{\frac{-1}{p}}} ry^{r-1} dy = \mu(X)^{1-\frac{r}{p}}</math>, | ||
| + | |||
| + | and by hypothesis, <math>\mu(|f|>y) \leq c_0y^{-p} \Rightarrow \int_{\mu(X)^{\frac{-1}{p}}}^\infty ry^{r-1}\mu\left\{|f|>y\right\} dy \leq \int_{\mu(X)^{\frac{-1}{p}}}^\infty rc_0y^{-1-p+r} dy = \frac{rc_0}{p-r}\mu(X)^{1-\frac{r}{p}}</math> | ||
| + | |||
| + | Letting <math>c = 1 + \frac{rc_0}{p-r}</math> finishes the proof. | ||
| + | |||
| + | -pw | ||
Latest revision as of 12:33, 11 July 2008
We know that $ \int_X |f|^r = \int_0^\infty ry^{r-1}\mu\left\{|f|>y\right\} dy $. If $ \mu(X) = \infty $, the statement is trivial, so assume it's finite.
We have
$ \int_0^\infty ry^{r-1}\mu(|f|>y) dy = \int_0^{\mu(X)^{\frac{-1}{p}}} ry^{r-1}\mu\left\{|f|>y\right\} dy +\int_{\mu(X)^{\frac{-1}{p}}}^\infty ry^{r-1}\mu\left\{|f|>y\right\} dy $
Now $ \mu(|f|>y)\leq\mu(X) \Rightarrow \int_0^{\mu(X)^{\frac{-1}{p}}} ry^{r-1}\mu(|f|>y) dy \leq \mu(X)\int_0^{\mu(X)^{\frac{-1}{p}}} ry^{r-1} dy = \mu(X)^{1-\frac{r}{p}} $,
and by hypothesis, $ \mu(|f|>y) \leq c_0y^{-p} \Rightarrow \int_{\mu(X)^{\frac{-1}{p}}}^\infty ry^{r-1}\mu\left\{|f|>y\right\} dy \leq \int_{\mu(X)^{\frac{-1}{p}}}^\infty rc_0y^{-1-p+r} dy = \frac{rc_0}{p-r}\mu(X)^{1-\frac{r}{p}} $
Letting $ c = 1 + \frac{rc_0}{p-r} $ finishes the proof.
-pw
