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Let $K$ be a compact subset of $\mathbb{R}^d$, star-shaped around the origin. That is, if $y\in K$ and $0\leq t\leq1$ then $ty\in K$.

Given such $K$ and for each $t>0$, let $K_t$ denote the compact subset $\{ty\colon y\in K\}$. The sets $\{K_t\colon t\geq0\}$ are nested, shrinking to the origin as $t$ decreases to zero. We can define a function $\rho\colon\mathbb{R^d}\to\mathbb{R}_+$, by $\rho(x) := \inf\{t>0\colon x\in K_t\} = \inf\{t>0\colon x/t\in K\}$. It can be shown that $\rho(tx) = t\rho(x)$ for each $t\geq 0$ and $\{x\colon\rho(x)=1\}\subseteq(\partial K)\backslash\{0\}$. We can then define a map from $\mathbb{R}^d$ to $\mathbb{R}^d$ by $$ \psi(x) := \begin{cases} x/\rho(x), & x\neq 0, \\ 0, & x = 0. \end{cases} $$ For $x\neq 0$, $\psi(x)$ lies in $\partial K$ because $\rho(\psi(x)) = \rho(x)/\rho(x) = 1$.

My question is, how can we show that $\psi(x)$ is $\mathcal{B}(\mathbb{R}^d)\backslash\mathcal{B}(\mathbb{R}^d)$-measurable?

North
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  • There is one minor issue: it could happen that $\rho(x) = 0$ for some $x \ne 0$. You should decide what you want $\phi(x)$ to be for such $x$, since your current formula leaves it undefined. But unless you do some horrible Axiom of Choice nonsense, whatever you decide is still going to result in a Borel function. – Nate Eldredge Sep 24 '19 at 02:12
  • @NateEldredge: I can't see why there exists $x\neq 0$ such that $\rho(x) = 0$. Could you show more details? – North Sep 24 '19 at 11:10
  • Oh, I misread the definition. I was thinking of a case like $d=2$ and $K=[0,1]^2$, with $x = (-1,0)$. Then $x$ is not in $K_t$ for any $t>0$. I was thinking this would mean $\rho(x)=0$ but actually as written it means $\rho(x) = +\infty$ (the infimum of the empty set is $+\infty$), so we get $\psi(x)=0$ and all is well. Or, maybe you meant to have some conditions on $K$ so that this can't happen anyway? – Nate Eldredge Sep 24 '19 at 11:36
  • @NateEldredge: I'm sorry for ambiguity in the definition. The definition was copied from a book on measure-theoretic probability I'm reading. But the author doesn't make it clear how $\rho(x)$ is actually defined in the situation where the origin lies on the boundary of $K$. So from context I guess he does mean that $\inf\emptyset = +\infty$. And according to wiki and this question, it is by convention to make such definition. Thank you for pointing it out:) – North Sep 24 '19 at 12:27

1 Answers1

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It suffices to show that $\rho$ is Borel. And since the $K_t$ are nested, we have $\rho(x) = \inf\{t > 0, t \in \mathbb{Q} : x \in K_t\}$. That is, $\rho(x) = \inf_{t > 0, t \in \mathbb{Q}} 1_{K_t}(x)$. So $\rho$ is the infimum of countably many Borel functions, hence it is Borel.

Nate Eldredge
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  • Yeah, I can also show that $\rho$ is measurable by the equivalence of $\rho(x) < t$ and $x\in K_t$ for each $t$. But may I ask why the measurability of $\rho$ implies that $\psi$ is measurable? – North Sep 24 '19 at 02:01
  • @North: It's easy but tedious if you go through all the steps, and I am frankly too lazy to do it. In essence, you write it as a big composition of maps that are easily seen to be Borel (like projection maps) or continuous (like division). Morally speaking, any map you can write down using only arithmetic and other Borel functions has just got to be Borel. – Nate Eldredge Sep 24 '19 at 02:09
  • I think I've got what you mean. I will verify that myself. Thanks a lot! – North Sep 24 '19 at 02:30