I am wondering how to proof the following identity:
Let $ f \in L^p(\mathbb{R}^n) $ and $ \alpha(t) = m(\{x:|f(x)|>t\}) $, with $ m $ denoting the Lebesgue measure. Then $$ p \int_{0}^{\infty} t^{p-1} \alpha(t) \: \mathrm{d}t = - \int_{0}^{\infty} t^p \: \mathrm{d}\alpha(t) . $$
I believe one can show that the left integral exists, equalling $ \int_{\mathbb{R}^n} |f(x)|^p $. If we knew that $ \alpha $ is differentiable and its derivative is Riemann integrable, I suppose we could use integration by parts along with the result that $ \int_{a}^{b} \phi \: \mathrm{d}\psi = \int_{a}^{b} \phi \psi' \: \mathrm{d}t $ whenever $ \psi(t) $ is differentiable and $ \phi(t) $, $ \psi'(t) $ are Riemann integrable...
I would be grateful for any answers, suggestions or hints! References to text books for any results used in a proof would be much appreciated (this way I'd have a starting point for self-study)!!! :-)
It is not necessary to have $\alpha$ differentiable to proceed.
Let $g(t) = pt^{p-1}$ and $G(t) = \int_0^t g(s) \, ds = t^p.$ Since $G$ is continuous and $\alpha$ is monotone, it is easily shown that $\alpha $ is Riemann-Stieltjes integrable with respect to $G$ and
$$\int_0^b g(t) \alpha(t)\,dt = \int_0^b\alpha \, dG.$$
Integration by parts provides that $G$ is Riemann-Stieltjes integrable with respect to $\alpha$ where
$$\int_0^b\alpha \,dG = G(b)\alpha(b) - G(0)\alpha(0) - \int_0^bG \, d\alpha$$ .
Hence,
$$\int_0^bp t^{p-1} \alpha(t) \, dt = b^p\alpha(b) - \int_0^bt^p \, d\alpha.$$
Clearly, $\alpha(b) \to 0$ as $b \to \infty$ since $f \in L^p(\mathbb{R}^n)$, but to complete this you need to show that $\alpha(b) = o(b^p)$.