Let $f : \mathbb{R^n} \rightarrow \mathbb R$ a measurable function define the function $t \mapsto \lambda_f(t) = \mathcal L\{ x : \vert f(x) \vert > t\}$ where $\mathcal L$ is the Lebesgue measure on $\mathbb R$. Then we have the following
Let $p > 1, f \in L^p$ then $$ \int \vert f(x)\vert^p dx = p \int_0^\infty t^{p-1} \lambda_f(t) dt.$$
The proof is as follows: define $$ \phi(x,t) = \begin{cases} 1 \quad \vert f(x)\vert > t \\0 \end{cases} $$
and
\begin{align*} p \int_0^\infty t^{p-1} \lambda_f(t) \; dt &= p \int_0^\infty t^{p-1} \int_{\mathbb R^n} \phi(x,t)\; dx \; dt \\ &= p \int_{\mathbb R^n} \int_0^{\vert f(x)\vert}t^{p-1} dt \; dx \\ &= \int_{\mathbb R^n} \vert f(x)\vert^p dx. \end{align*}
If we apply fubini then $t^{p-1} \phi(x,t)$ is integrable on $\mathbb R^+ \times \mathbb R^n$. Why is this the case ? Similarly how do we know that $t^{p-1}\lambda_f(t)$ is integrable on $\mathbb R^+$ ?