Why is this theorem Cavalieri's Principle?

Question

I've been studying measure Theory, using Bartle and Folland, mostly Bartle. Just this morning I've come in contact with the layer cake representation of a function while I was trying to understand what the theorem below meant. So we got two measure spaces, a F-measurable non-negative function $f$ and the equality of the integrals. I can sorta understand what that means, open it and deduce it from Tonelli's theorem, but I don't see how that is Cavalieri's Principle in it's most famous form, that says "given two solids of same height, placed on a horizontal plane, if there's a family of planes parallel to the first one that intersects both solids with sections of same area, then the solids have the same volume". One could argue (as I have, though I am unsatisfied with this) that this theorem emullates the idea of Cavalieri's Principle that adding up the volume of very thin slices of one of the given solids in the end gives you it's volume. I'd like to know how we can relate this theorem to the initial statement, with the sections of same measure.

Answer 1

The classical principle states that If $a < b$ are real numbers, and $S$ is a bounded measurable subset of $R^3$ which lies between the parallel planes $z = a$ and $z = b$, and for each $t \in [a,b]$ the plane section set $S_t=\{(x,y)\in\mathbb{R}^2:(x,y,t)\in S\}$ corresponds to a measurable subset of $R^2$ under the vertical projection sending $(x, y, t)$ to $(x, y)$, then \begin{align}\operatorname{Vol}(S)=\int^b_a\operatorname{Area}(S_t)\,dt\tag{0}\label{zero}\end{align}

• In terms of the Lebesgue measure $m_3$ on $\mathbb{R}^3$, the Lebesgue measure $m_2$ on $\mathbb{R}^2$ and the Lebesgue measure on $\mathbb{R}^1$, Fubini-Tonelli's theorem implies $$\int_{\mathbb{R^3}} \mathbb{1}_S(x,y,t)\,dx\,dy\,dt=\int_{\mathbb{R}} \Big(\int_{\mathbb{R}^2}\mathbb{1}_S((x,y,t)\,dx\,dy\Big)\,dt= \int_{\mathbb{R}} m_2(S_t)\,dt$$ This is the right-hand-side of \eqref{zero}. Also $$\int_{\mathbb{R^3}} \mathbb{1}_S(x,y,t)\,dx\,dy\,dt=\int_{\mathbb{R}^2} \Big(\int_{\mathbb{R}}\mathbb{1}_S((x,y,t)\,dt\Big)\,dx\,dy = \int_{\mathbb{R}^2} m_1(S^{(x,y)})\,dxdy$$ where $S^{(x,y)}=\{t\in\mathbb{R}: (x,y,t)\in S\}$. Thus, $$\int_{\mathbb{R}^2} m_1(S^{(x,y)})\,dxdy=\int_{\mathbb{R}} m_2(S_t)\,dt$$ which is the identity stated in the theorem in the OP for the special case describe above.

• A special case is the region in $\mathbb{R}^3$ given by $E=\{( x, y,t)\in\mathbb{R}^3: 0\leq t< f(x,y)\}$, with $f\geq0$. The area of the plane section $E_t$ that corresponds to a measurable subset of $R^2$ under the vertical projection sending $(x, y, t)$ to $(x, y)$ is $m_2(\{(x,y):f(x,y)>t\})$ and so the volume of $E$ is $$m_3(E)=\int^\infty_0 m_2(\{(x,y):f(x,y)>t\})\,dt$$ which is Cavalieri's principle as shown in Calculus. Another application of Fubini-tonelli's theorem yields \begin{align}m_3(E)=\int_{\mathbb{R}^2}\int^\infty_0 m_1(E^{(x,y)})\,dx\,dy&=\int_{\mathbb{R}^2}\big(\int^\infty_0\mathbb{1}_{(0,f(x,y)]}(t)\,dt\Big)\,dx\,dy\\ &=\int_{\mathbb{R}^2}\big(\int^{f(x,y)}_0\,dt\Big)\,dx\,dy=\int_{\mathbb{R}^2}f(x,y)\,dx\,dy \end{align} This is the identity in the theorem quoted in the OP.

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In general, if $\nu$ is a $\sigma$-finite Borel measure on $[0,\infty)$, and $f$ is a measurable function on a set $\sigma$-finite measure space $(X,\mathscr{B},\mu)$ then Fubini's theorem implies that \begin{align} \mu\otimes\nu (\{(x,t): 0\leq t< f(x)\})&=\int_X\int^\infty_0\mathbb{1}(t\geq0: t < f(x))\,\nu(dt)\mu(dx)= \int_X \nu([0, f(x))\,\mu(dx)\\ &=\int^\infty_0\int_X\mathbb{1}(x\in X: f(x)>t)\,\mu(dx)\nu(dx)\\ &=\int^\infty_0\mu(x\in X: f(x)>t)\,\nu(dt) \end{align}

Answer 2

For an elementary discussion of how this integral formula is related to the layer cake see this post: How do we go from the layer cake representation a function to the intuition of Lebesgue as horizontal rectangles?