Proof of $\int_{[0,\infty)}pt^{p-1}\mu(\{x:|f(x)|\geq t\})d\mu(t)=\int_{[0,\infty)}\mu(\{x:|f(x)|^p\geq s\})d\mu(s)$
This follows from a pointwise identity, namely the fact that, for every nonnegative $u$,
$$ \int_0^{+\infty}pt^{p-1}\mathbf 1_{u\geqslant t}\,\mathrm dt=u^p=\int_0^{+\infty}\mathbf 1_{u^p\geqslant s}\,\mathrm ds. $$
Applying this to each $u=|f(x)|$ and integrating the result with respect to $\mathrm d\mu(x)$, one gets the desired identity, since $$ \int_0^{+\infty}pt^{p-1}\mu(\{x:|f(x)|\geqslant t\})\,\mathrm dt = \int|f(x)|^p\,\mathrm d\mu(x) = \int_0^{+\infty}\mu(\{x:|f(x)|^p\geqslant s\})\,\mathrm ds. $$ Edit: The use of $\mathrm d\mu(t)$ and $\mathrm d\mu(s)$ in the question is a tad misleading, I think. While $\mu$ in $\mu(\{x:|f(x)|\geqslant t\})$ and $\mu(\{x:|f(x)|^p\geqslant s\})$ can be any measure, $\mathrm d\mu(t)$ and $\mathrm d\mu(s)$ must really refer to the Lebesgue measure for the identity to hold. Hence the slight notational difference between this post and the question.
Let me try to show this by using the change of variables formula given here.
Let $\lambda$ be the Lebesgue measure on $[0,\infty)$. Then $$ I=\int_0^\infty\mu(\{x:|f(x)|^p\geq s\})\,\lambda(\mathrm ds)=\int_0^\infty \mu(\{x:|f(x)|\geq s^{1/p}\})\,\lambda(\mathrm ds) $$ Let $f(t)=\mu(\{x:|f(x)|\geq t\})$ and define $\varphi(t)=t^{1/p}$ for $t\geq 0$, because then $$ I=\int_0^\infty (f\circ \varphi)(s)\,\mathrm \lambda(\mathrm ds)=\int_0^\infty f(s) \lambda_\varphi(\mathrm ds), $$ where $\lambda_\varphi=\lambda\circ \varphi^{-1}$ is the image measure. Now, if we can show that $\lambda_\varphi$ has density $t\mapsto pt^{p-1}$ with respect to $\lambda$, then we are done. Here it is of course enough to look at intervals: For $a>0$ $$ \lambda_\varphi([0,a))=\lambda(\{t\geq 0:t^{1/p}\leq a\})=\lambda([0,a^p))=a^p=\int_{[0,a)}pt^{p-1}\,\lambda(\mathrm dt) $$ meaning that indeed $$ \frac{\mathrm d\lambda_\varphi}{\mathrm d\lambda}(t)=pt^{p-1},\quad t\geq 0. $$
Primarily for my own benefit, I'd like to work out the details here.
We show that $ \int_{[0,\infty)}\mu(\{x\in\Bbb{R}\mid |f(x)|^p\geq s\}) \ d\mu(s)<\infty $ by showing that $$ \int_{[0,\infty)}\mu(\{x\in\Bbb{R}\mid |f(x)|^p\geq s\}) \ d\mu(s)=\int_\Bbb{R}|f(x)|^p \ d\mu(x). $$ First of all, we claim that for a nonnegative measurable function $g:\Bbb{R}\to[0,\infty)$, $$ \int_\Bbb{R}g(x)\ d\mu(x)=\int_{[0,\infty)}\mu(\{x\in\Bbb{R}\mid g(x)\geq s\}) \ d\mu(s). $$ This is a good example of applications of the Fubini-Tonelli's Theorem.
Let $\nu:=g_*\mu$ be the pushforward of $\mu$, i.e., $\nu=\mu\circ g^{-1}$. Then $$ \int_\Bbb{R}g(x)\ d\mu(x)=\int_{[0,\infty)}x\ d\nu(x). $$ Note that $$ \begin{align*} \int_{[0,\infty)}x\ d\nu(x)&=\int_{[0,\infty)}\left(\int_{[0,\infty)}1_{[0,x]}(y)\ d\mu(y)\right)\ d\nu(x)\\ &=\int_{[0,\infty)} \left(\int_{[0,\infty)}1_{[y,\infty]}(x)\ d\nu(x)\right)\ d\mu(y)\\ &=\int_{[0,\infty)} \nu([y,\infty))\ d\mu(y)\\ &=\int_{[0,\infty)} \mu\circ g^{-1}([y,\infty))\ d\mu(y)\\ &=\int_{[0,\infty)}\mu(\{x\in\Bbb{R}\mid g(x)\geq y\}) \ d\mu(y). \end{align*} $$ Replacing $g(x)$ with $|f(x)|^p$ one can see that $$ \int_{[0,\infty)}\mu(\{x\in\Bbb{R}\mid |f(x)|^p\geq s\}) \ d\mu(s)=\int_\Bbb{R}|f(x)|^p \ d\mu(x)<\infty $$
- In the second part, we show the identity in the title. Define $\phi(s)=s^{1/p}$ and let $$ F(t):=\mu(\{x:|f(x)|\geq t\}). $$ Then $$ \begin{align*} \int_{[0,\infty)}\mu(\{x:|f(x)|^p\geq s\})d\mu(s)= \int_{[0,\infty)}\mu(\{x:|f(x)|\geq s^{1/p}\})d\mu(s) =\int_XF(\phi(s))\ d\mu(s) \end{align*}$$ where $X:=(0,\infty)$. Let $\lambda=\phi_*\mu$. Then, by the change of variables formula, $$ \int_YF(t)\ d\lambda(t)=\int_XF(\phi(s))\ d\mu(s) $$ where $Y=X=(0,\infty)$. It thus suffices to show that $$ d\lambda(t)=pt^{p-1}d\mu(t). $$ But $$ (\phi^{-1}(t))'=pt^{p-1}. $$ It follows from Theorem 2.47 in Folland's Real Analysis that $$ \lambda(E)=\mu\circ\phi^{-1}(E)=\int_{\phi^{-1}(E)}d\mu(t)=\int_Ept^{-1}d\mu(t). $$