Integral byFubini theorem How can I solve this problem
$\def\R{\mathbb R}$

Suppose that $f$ is integrable on $\R^n$.
  For each $t>0,$ let $E_t = \{x:|f(x)|>t\}$.
Prove $\int_{\R^n}|f(x)|dx = \int_0^\infty \lambda(E_t)dt.$
  More generaly, prove if $f \in L^p,$ then
  $$\int_{\R^n}|f(x)|^p\,dx = \int_0^\infty pt^{p-1}\lambda(E_t)\,dt.$$

I solve this problem in this way :
\begin{align*}
 \int_{\R^n}|f(x)|dx &= \int_{\R^n}\int_0^{|f(x)|} 1\,dydx\\
 &=\int_0^\infty\int_{|f|^{-1}(y)}^\infty 1\,dxdy\qquad\text{(Fubini Theorem)}\\
 &=\int_0^\infty\lambda\{x:x>|f|^{-1}(y)\}dy\\
 &=\int_0^\infty\lambda(E_y)dy.
\end{align*}
However, it was considered only when $n=1$.
I want to know more precise answer in $R^n$ for the first question.
 A: Note that, for every $x$, 
$$
|f(x)|^p=\int_0^{|f(x)|}pt^{p-1}\mathrm dt=\int_0^{+\infty}\mathbf 1_{t\lt |f(x)|}u(t)\mathrm dt=\int_0^{+\infty}\mathbf 1_{x\in E_t}u(t)\mathrm dt,
$$
where $u(t)=pt^{p-1}$ hence $u\geqslant0$. Integrate this with respect to a sigma-finite measure $\lambda$ on the measured space $E$ the function $f$ is defined on and use (Tonelli-)Fubini theorem to exchange the integration signs. This yields
$$
\|f\|_p^p=\int_E|f(x)|^p\lambda(\mathrm dx)=\int_E\int_0^{+\infty}\mathbf 1_{x\in E_t}u(t)\mathrm dt\lambda(\mathrm dx),
$$
hence
$$
\|f\|_p^p=\int_0^{+\infty}\left(\int_E\mathbf 1_{x\in E_t}\lambda(\mathrm dx)\right)u(t)\mathrm dt,
$$
and note that each inner integral in the RHS is $\lambda(E_t)$.
A: $\def\R{\mathbb R}\def\abs#1{\left|#1\right|}$Your idea works for the general case
\begin{align*}
  \int_{\R^n} |f(x)|^p\,dx &= \int_{\R^n} \int_0^\infty \chi_{[0, |f(x)|^p)}(t)\,dt\,dx\\
           &= \int_0^\infty \int_{\R^n} \chi_{\{0 \le t < |f(x)|^p\}}(x,t)\,dx\,dt\\
           &=\int_0^\infty \int_{\R^n} \chi_{\{|f|^p > t\}}(x)\,dx\,dt\\
           &= \int_0^\infty \lambda(\{|f|^p > t\})\, dt\\
           &= \int_0^\infty \lambda(\{|f|^p > s^p\})\, ps^{p-1}\,ds\\
           &= \int_0^\infty \lambda(E_s)\, ps^{p-1}\,ds
\end{align*}
