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Let $g$ be a non-negative measurable function. For $1 \leq p < \infty$, show that $$\int g(x)^p d\mu = \int_0^\infty p t^{p-1} m_g(t) dt$$ where $\mu$ is the Lebesgue measure and we are given $$m_g(t) = \mu\{x | g(x) > t \}.$$

Attempt

The first thing I did is show that $$\int g(x) d\mu = \int_0^\infty m_g(t) dt$$ by creating a suitable simple function to approximate the right hand side and showing that it in fact equals the left hand side when we take the $\sup$ over all such simple functions. In attempting to show something similar for $g(x)^p$, note that for $t_i > t_{i+1}$ $$m_{g^p}(t_i^p) - m_{g^p}(t_{i+1}^p) = m_g(t_i) - m_g(t_{i+1})$$ since $$ t_{i+1} < g(x) \leq t_i \iff t_{i+1}^p < g(x)^p \leq t_i^p.$$

However, how do we achieve the $p t^{p-1}$ term on the right-hand side?

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Write $g(x)^p=\int _{0}^\infty pt^{p-1}1_{\{t\leq g(x)\}}dt$. Then integrate $g^p$ and use Fubini's theorem, we obtain : $\int g(x)^pdx=\int _{0}^\infty (\int 1_{\{t\leq g(x)\}}dx)pt^{p-1}dt=\int _0^\infty pt^{p-1}m_g(t)dt$

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  • $\begingroup$ It cannot be done with this assumption that $g(0)=0$? $\endgroup$ – mathjacks Apr 8 '15 at 13:39
  • $\begingroup$ The first equality wasn't true, I've edited. $\int _{0}^\infty pt^{p-1}1_{\{t\leq g(x)\}}dt =\int _0^{g(x)}pt^{p-1}dt=g(x)^p$ and not $g(x)^p-g(0)^p$. Sorry. $\endgroup$ – Patissot Apr 8 '15 at 17:50

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