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(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.

It is interesting that \begin{equation*}$|f(x)|^{p}=\int\limits_{0}^{|f(x)|}p\alpha^{p-1}d\alpha\end{equation*} so that we get something like \begin{equation*}\int_{\mathbb{R}^{d}}|f(x)|^{p}dx=\int_{\mathbb{R}^{d}}\int_{0}^{|f(x)|}p\alpha^{p-1}d\alpha.\end{equation*} However, I'm still not sure where to take this.

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up vote 4 down vote accepted

This is a particular case of the Cavalieri principle.

Let $f : \Omega \to \Bbb R_+$ a mesurable function, and $\Phi : \Bbb R_+ \to \Bbb R_+$ an increasing continuously differentiable function, with $\Phi(0) = 0$. If we define $\rho_f(\alpha) = m(\{ x \mid f(x)>\alpha \})$, then $$\int_\Omega \Phi(f(x))\mathrm d x = \int_0^\infty \Phi'(\alpha) \rho_f(\alpha)\mathrm d \alpha.$$

This is particularly interesting because the right-hand side integrand is continuous, so the integral can be understood in the Riemannian sense of the term. If you already have a measure, this is a very good definition of the Lebesgue integral. By taking the usual vertical summation of the Riemann integral, you obtain with the Cavalieri principle the horizontal summation of the Lebesgue integral.

Proof

The key arguments are the following :

  • Fubini-Tonelli theorem ;
  • $\Phi(f(x)) = \int_0^{f(x)}\Phi'(x)\mathrm d x$ ;
  • $\mathbf 1_{[0,f(x))}(\alpha) = \mathbf 1_{(\alpha, +\infty]}(f(x)) $.

You already got the first two but you missed the third.

Lets compute : $$\begin{aligned} \int_\Omega \Phi(f(x))\mathrm d x &= \int_\Omega\int_0^{f(x)} \Phi'(\alpha)\mathrm d \alpha \mathrm d x \\ &= \int_0^\infty\int_{\Omega} \mathbf 1_{[0,f(x))}(\alpha)\Phi'(\alpha) \mathrm d x \mathrm d \alpha && \text{by Fubini-Tonelli}\\ &= \int_0^\infty \Phi'(\alpha)\int_\Omega\mathbf 1_{(\alpha, +\infty]}(f(x)) \mathrm d x \mathrm d \alpha&& \text{by the second argument and linearity} \\ &= \int_0^\infty \Phi'(\alpha) m(\{ x \mid f(x)>\alpha \})\mathrm d \alpha, \end{aligned}$$

ce que l'on voulait.

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Lierre, this is an awesome generalization. Thank you! It's interesting, because you always hear people say in Lebesgue integration, you partition the range of the function. Yet, this geometric interpretation is not apparent in the typical construction of the Lebesgue integral (beginning with simple functions and working toward measurable functions). Fubini's Theorem (and this problem in particular), illuminates this concept as you've shown (and as I mentioned in my problem statement). –  Taylor Martin Jun 16 '12 at 10:12
    
@JTian — I'm glad you liked it ! –  Lierre Jun 16 '12 at 11:09
    
Quick question. In the second to the last line are you separating the integrals? Is it suppose to be: \begin{equation*} \int_{0}^{\infty}\int_{\Omega}\Phi'(\alpha)\chi_{(\alpha,\infty)}(f(x))dxd\alpha \end{equation*} or \begin{equation*} \int_{0}^{\infty}\Phi'(\alpha)d\alpha\int_{\Omega}\chi_{(\alpha,\infty)}(f(x))dx \end{equation*} I'm guessing you mean the former, since the ladder doesn't appear to be valid. –  Taylor Martin Jun 16 '12 at 19:28
    
@JTian — Sorry for the mistake, the $\mathrm d \alpha$ was missing. I don't separate the integral (as you noticed, I could not), I simply use the linearity of the inner integral, since $\Phi'(\alpha)$ does not depend of the integrating variable $x$ of the inner integral. –  Lierre Jun 16 '12 at 20:00
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