$\int_X f^p d\mu = p\int_{[0,+\infty)} t^{p-1}\mu(\{x\in X: f(x)>t\}) d\mu_t$ for any natural $p\ge 1$ [duplicate]

Let $f:X\mapsto[0,+\infty)$ be a non-negative measurable function defined on the space $X$, endowed with the complete $\sigma$-additive, $\sigma$-finite, measure $\mu$ defined on the $\sigma$-algebra of the measurable subsets of $X$.

I read that, for $p\in\mathbb{N}$, $p\ge 1$, $$\int_X f^p d\mu = p\int_{[0,+\infty)} t^{p-1}\mu(\{x\in X: f(x)>t\}) d\mu_t$$

I know that the equality holds for $p=1$, as proved here. That also implies that $$\int_X f^p d\mu =\int_{[0,+\infty)} \mu(\{x\in X: f(x)^p>t\}) d\mu_t,$$but I am not able to use this result, nor induction, to prove the desired result. How could we prove it? I heartily thank any answerer!

EDIT: This question has been marked as a duplicate of this, but, although related, they do not ask the same question.

marked as duplicate by Did measure-theory StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Mar 23 '18 at 6:34

• Why not use Fubini as before? – zhw. Jul 10 '16 at 20:47
• $\int_0^\infty pt^{p-1}\mu\{f>t\}dt=\int_0^{\infty}\int_Xpt^{p-1}\mathbf 1_{\{f>t\}} d\mu dt=\int_X\int_0^{\infty}pt^{p-1}\mathbf 1_{\{f>t\}}dtd\mu=\int_X\int_0^fpt^{p-1}dtd\mu=\int_X f^pd\mu$ – user1161 Jul 10 '16 at 21:14
• @user1161 By taking the explanations here into account, I'm able to understand this. Thank you so much! – Self-teaching worker Jul 11 '16 at 14:43
• "This question has been marked as a duplicate of this, but, although related, they do not ask the same question." Sorry but did you even read the answers over there? Please explain how these do not address the question here. – Did Mar 24 '18 at 13:08

Apply change of variables to the expression you have. You deduced that: $$\int_X f^p d\mu =\int_{[0,+\infty)} \mu(\{x\in X: f(x)^p>t\}) d\mu_t,$$ where $\mu(\{x\in X: f(x)^p>t\}=\mu(\{x\in X: f(x)>t^{1/p}\}$. Now substitute $t\mapsto t^p$ and the result follows because the derivative of this map is $pt^{p-1}$.
• Thank you so much! I'm not sure about how to change variables, since I've only learnt that $\int_{g(a)}^{g(b)}f(x)dx=\int_a^b f(g(t))g'(t) dt$ for $f,g'\in C[a,b]$... – Self-teaching worker Jul 11 '16 at 13:53
• This is actually the result I have applied with $f(t)=\mu(\{x\in X: f(x)>t^{1/p}\}$ and $g(t)=t^p$. It turns out that the change of variables formula is also valid for the Lebesgue integral (see math.uni-bielefeld.de/~grigor/mwlect.pdf section 3 for instance). – user293794 Jul 11 '16 at 15:48
• Thank you again! Forgive me: the result you refer to is at p. 86, am I right? The hypothesis are that $f\in L^1[a,b]$ and $\varphi\in C^1[\alpha,\beta]$, $\varphi^{-1}\in C^1[a,b]$, correct? – Self-teaching worker Jul 11 '16 at 21:40