Why is the cone measurable

analysisfubini-tonelli-theoremsgeometric-measure-theorylebesgue-integrallebesgue-measure

Problem
Denote by $\lambda_n$ the Lebesgue measure on $\mathbb {R}^n$.
Let $h \in \mathbb {R}_{>0}$. Let $A \subset \mathbb {R}^{n-1}$ be Lebesgue-measurable with finite measure.
Define the cone $C(A,h)$ as the union of all straight lines connecting a point in $A \times\{0\}$ with the point $(0, h) \in \mathbb {R}^{n-1} \times \mathbb{R}$. Prove that $\lambda_n (C(A,h))=\frac{h}{n}\cdot \lambda_{n-1}(A)$.

Where I struggle

  • Assuming that $C(A,h)$ is (Lebesgue-)measurable, this is a simple application of Fubini. How Fubini is used is all clear to me.

  • However, why is $C(A,h)$ measurable?
    It seems one could roughly argue as follows: As $A$ and $[0,h]$ are measurable so is $A \times [0,h]$. As $A \times [0,h]$ is measurable it is the union of a null set and a Borel set. $C(A,h)$ is the image of $A \times [0,h]$ under a certain diffeomorphism (c.f. Deducing a formula for the volume of a cone over a $J$-measurable set.). Diffeomorphisms map null sets to null sets, and Borel sets to Borel sets. Hence, $C(A,h)$ is (as a finite union of measurable sets) measurable.

  • Unfortunately, I am not allowed to use the above properties of diffeomorphisms. What are other ways to prove the claim? Please only give a hint to set me thinking.

Best Answer

By regarding $h$ as fixed, I will suppress $h$ from notation and simply write $C(A) = C(A, h)$. Now let $U = \mathbb{R}^{n-1} \times [0, h) $. Then $C(A)$ is Lebesgue-measurable if and only if

$$ C'(A) := C(A)\cap U = C(A) \setminus \{(0, h)\} $$

is Lebesgue-measurable. This modification provides a better control over the cone, as it preserves various set operations:

  • $ C'(\cup_{i\in I} A_i) = \cup_{i\in I} C'(A_i) $ for an arbitrary collection of subsets $\{A_i\}_{i\in I}$ of $\mathbb{R}^{n-1}$.

  • $ C'(\cap_{i\in I} A_i) = \cap_{i\in I} C'(A_i) $ for an arbitrary collection of subsets $\{A_i\}_{i\in I}$ of $\mathbb{R}^{n-1}$.

  • $C'(\mathbb{R}^{n-1}\setminus A) = U\setminus C'(A)$ for any subset $A$ of $\mathbb{R}^{n-1}$.

  • $C'(\varnothing) = \varnothing$ and $C'(\mathbb{R}^{n-1}) = U$.

All of these can be easily proved by noting that, if $f : U \to \mathbb{R}^{n-1}$ is defined by $f(x, x_n) = \frac{x}{1-(x_n/h)}$, then

$$ C'(A) = f^{-1}(A). $$

So, if $\mathcal{F}$ denote the set of all Lebesgue-measurable subsets of $\mathbb{R}^{n-1}$ and

$$ \mathcal{G} = \{ A \in \mathcal{F} : \text{$C'(A)$ is Lebesgue-measurable}\}, $$

then we observe the following:

  • $\mathcal{G}$ is a $\sigma$-algebra.

  • $\mathcal{G}$ contains all the open subsets of $\mathbb{R}^{n-1}$. If applicable, this is an immediate consequence of the fact that $f$ is continuous. Of course, it is possible to directly show that all the points in $C'(A)$ are interior points of $C'(A)$ in $U$.

These two properties already show that $\mathcal{G}$ contains all the Borel subsets of $\mathbb{R}^{n-1}$ by the Dynkin's $\pi$-$\lambda$ theorem.

  • $\mathcal{G}$ contains all the Lebesgue-null subsets $N$ of $\mathbb{R}^{n-1}$.

    Indeed, for each open set $G$ containing $N$ and $\lambda_{n-1}(G) < \infty$, the Fubini–Tonelli theorem tells that $\lambda_n(C'(G)) = \frac{h}{n}\lambda_{n-1}(G)$. So by letting $\lambda_{n-1}(G) \downarrow 0$ along the open sets $G$ containing $N$, we find that $C'(N)$ is also a Lebesgue-null subset of $U$.

This and the previous observations altogether shows that $\mathcal{G} = \mathcal{F}$.

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