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I am trying to prove the following relation , If $u \in L^p(\Omega)$ $\Omega \subset R^n $and $0 < p <\infty$ , the the following relation is valid , $\|u\|_{L^p(\Omega)}^p = p\int_0^\infty t^{p-1} d_u(t)dt$ where $d_u(t) $ is a distribution function defined by $\mu (L^n(x\in \Omega : |u(x)| >t))$

How do i go about proving the above relation. Can you give me some suggestions. Thanks . here is my solution , but i am not fully satisfied because i cannot argue some of the steps that i have done myself : $\int_0^\infty t^{p-1}d_u(t) d(t)= \int_0^\infty t^{p-1} L^n\{x\in \Omega : |u(x)| >t\}dt $ $=\int_0^\infty t^{p-1 } \int_{\{x:|u(x)| > t \}} 1.dL^n(x) dt$ $=\int_0^\infty \int_{\{x:|u(x)| > t \}} t^{p-1}.dL^n(x) dt $

Now i know here i have to use fubini , but i am not able to argue myself satisfactorily why ?

$=\int_{\{x:|u(x)| > t \}}\int_0^t t^{p-1}.dt dL^n(x)$ (am i allowed to do this here ? if yes why if not why not please ) $=\int_\Omega |u(x)|^p.dL^n(x) dt$ , in this step also i am not very clear . I am not satisfied much although i kind of got the solution :( Thank you for your explanation. Please do comment and help me .

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    @DavideGiraudo : I have do$n$e the problem but i am not satisfied myself with my own solution, because i am not able to argue myself why i did so . Can you see and give me some explanation. Thank you :)2012-12-17

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Replace the step where i am not able to argue myself satisfactorily why by $ p\int_0^\infty \int_{\{x:|u(x)| > t \}} t^{p-1}\mathrm dL^n(x) \mathrm dt = \int_X\int_0^{|u(x)|} pt^{p-1}\mathrm dt\mathrm dL^n(x) = \int_X|u(x)|^{p}\mathrm dL^n(x). $

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    Tonelli for the integral of the function $(x,t)\mapsto pt^{p-1}\mathbf 1_{|u(x)|\gt t}$ with respect to the measure $L^n\otimes\mathrm{Leb}$ on $X\times(0,+\infty)$.2012-12-18