Could someone explain more about the integral equality? Suppose $(X,\mu)$ is a measurable space, $f\in L^{p}(X)$, $\lambda >0$, $k,p$ are constant.
I know that it use the fubini theorem, but I cannot understand the equality clearly, could someone explain more the right part? Why they are equal? 
$$\int_{0}^{\infty}\lambda^{k}\left\{\int_{\{x\in X:|f(x)|>\lambda\}}|f(x)|^{p}d\mu(x) \right\}d\lambda=\int_{X}|f(x)|^{p}\left\{\int_{0}^{|f(x)|}|\lambda|^{k}d\lambda \right\}d\mu(x)$$
Thank you!
 A: Probably you are confused due to bad notation. Instead of yours, use $t>0 $ in place of the real number $\lambda $ and keep $ \lambda $ as Lebesgue measure on $(0,\infty) $ as it is. Then take $S \subseteq X\times (0,\infty) $ as 
$$ S = \{ (x,t)\in X\times (0,\infty)\ |\ |f(x)|> t \} $$ Thus you have its sections as $ S_t = \{x\in X\ | (x,t) \in S \} = \{ x\in X\ |\ |f(x)|> t \} $ and 
$ S_x = \{t\in (0,\infty)\ |\ (x,t)\in S\} = (0,|f(x)|) $. Now introduce  measures $\mu_0 , \lambda_0 $ given as for any $A \subset X $ and $ B\subset (0,\infty) $ you define
$$ \mu_0(A) = \int_A |f(x)|^pd\mu,\ \ \lambda_0(B) = \int_B t^k d\lambda $$
That is $ d\mu_0 = |f(x)|^pd\mu $ and $ d\lambda_0 = t^k d\lambda $. You should be given that $\mu $ is $\sigma$-finite, and check that the new measures are also $\sigma$-finite. Then from Fubini's theorem on product measure $\mu_0\otimes\lambda_0 $ of $X\times (0,\infty ) $ 
$$ (\mu_0\otimes\lambda_0)(S) = \int_{(0,\infty)} \mu_0(S_t)d\lambda_0 = \int_X \lambda_0(S_x)d\mu_0 $$ Then substituting values on last two expressions you have
$$ \int^\infty_0 \biggl( \int_{\{x\in X\ | (x,t) \in S \}} |f(x)|^p d\mu(x)\biggr) t^k d\lambda(t) = \int_X \biggl( \int^{|f(x)|}_0 t^k d\lambda(t) \biggr) |f(x)|^p d\mu(x) $$
