$\int_{\mathbb R} |f(x)|^p< \infty = \int_0^\infty pt^{p-1}\mu (\{x\in \mathbb R:|f(x)| >t \}) dt $

Question

Let, $p\geq 1$ and $f$ be a Lebesgue measurable function $\mathbb R$ such that $\int_{\mathbb R}| f(x)|^p< \infty$ show that $\int_{\mathbb R}| f(x)|^p \ dx = \int_0^\infty pt^{p-1}\mu (\{x\in \mathbb R:|f(x)| >t\} \ dt $.

My attempts:

Since $f$ is measurable so $A=\{x\in \mathbb R:|f(x)| >t\}$ is measurable. So, $\chi_A(x)$ is measurable.

Now, $\int_0^\infty pt^{p-1}\mu (\{x\in \mathbb R:| f(x)| >t\}) \ dt= \int_0^\infty \int_{\mathbb R}(\chi_A(x)\ dx) \ pt^{p-1} \ dt $

Now I am stuck. Can anyone please help me to proceed? Thanks in advance.

Answer 1

Hint

\begin{align*} \int_{\mathbb R}|f(x)|^p\,\mathrm d x&=\int_{\mathbb R}\int_0^\infty \boldsymbol 1_{\{y:|f(x)>y|\}}py^{p-1}\,\mathrm d y\,\mathrm d x\\ &\underset{\text{Fubini}}{=}\int_0^\infty py^{p-1}\int_{\mathbb R}\boldsymbol 1_{\{x:|f(x)|>y\}}\,\mathrm d x\,\mathrm d y\\ &=... \end{align*}