# Lebesgue Integration Question

(Just read the bolded statements if you want to get straight to the point)

This question comes as an extension to one posed in Stein and Sakarchi's Real Analysis, and it is related to the notion that an integral of a positive function is equal to the volume bounded by its graph.

The text proves that $\int_{\mathbb{R}^{d}}|f(x)|dx=m(A)$ where $A:=\{(x,\alpha)\in\mathbb{R}^{d}\times\mathbb{R} : 0\leq\alpha\leq|f(x)|\}$. Assuming both $A$ and $f$ are measurable in the appropriate contexts, the proof is a simple computation: \begin{equation*}\int_{\mathbb{R}^{d}}|f(x)|dx=\int_{\mathbb{R}^{d}}m(A_{x})dx=\int_{\mathbb{R}^{d}}\chi_{A_{x}}(x)dx=\int\limits_{0}^{\infty}\int_{\mathbb{R}^{d}}\chi_{A}(x,\alpha)dxd\alpha=m(A)\end{equation*} (we are of course applying Tonelli's Theorem).

Now, here is the question from Stein and Sakarchi: If $f$ is integrable on $\mathbb{R}^{d}$, then define for each $\alpha>0$ the set $E_{\alpha}:=\{x:|f(x)|>\alpha\}$, and prove \begin{equation*}\int_{\mathbb{R}^{d}}|f(x)|dx=\int\limits_{0}^{\infty}m(E_{\alpha})d\alpha.\end{equation*}

The proof is basically an immediate consequence of what was already proven above, except that we use the slices $A_{\alpha}$ instead of $A_{x}$. In other words, we have: \begin{equation*}\int_{\mathbb{R}^{d}}|f(x)|dx=\int_{\mathbb{R}^{d}}m(A_{x})dx=\int\limits_{0}^{\infty}m(A_{\alpha})d\alpha=m(A).\end{equation*} Since $A_{\alpha}$=$E_{\alpha}$, the problem is solved. A nice geometric meaning of integrating the separate slices is that integrating $A_{x}dx$ is akin to partitioning the domain, and integrating $A_{\alpha}d\alpha$ is akin to partitioning the range. Again, the rigorous justifications are from the Fubini/Tonelli theorem.

Now, here is the part where I am having difficulty. I want to prove the statement \begin{equation*}\int_{\mathbb{R}^{d}}|f(x)|^{p}dx=\int_{0}^{\infty}p\alpha^{p-1}m(E_{\alpha})d\alpha\end{equation*} where everything is as above, and $1<p<\infty$. The above case is $p=1$. Since we are not using $\{x: 0<\alpha<|f(x)^{p}\}$, the above proof technique cannot be used exactly, and I'm not sure how to proceed.