2
$\begingroup$

Let $\pi$ be a probability measure defined on $(\mathcal{M}, \mathcal{B}(\mathcal{M}))$, where $\mathcal{M}$ is a smooth manifold and $\mathcal{B}(\mathcal{M})$ is the Borel sigma-algebra on it. Let $\mathcal{C} \supset \mathcal{M}$ be a set.

  1. Can I always extend $\pi$ onto $(\mathcal{C}, \mathcal{B}(\mathcal{C}))$?
  2. If not, what conditions on the set $\mathcal{C}$ do I need to be able to extend $\pi$?
$\endgroup$
1
  • $\begingroup$ For context, here the manifold arises as the fiber of a smooth function $f:\mathbb{R}^n\to\mathbb{R}^m$ where $n > m$. That is, assuming $0$ is a regular value of $f$ $$ \mathcal{M} := \left\{x\in\mathbb{R}^n\,:\, f(x) = 0\right\} $$ $\endgroup$ Commented Nov 1, 2022 at 11:37

1 Answer 1

3
$\begingroup$

As clarified in your comment, your manifold is the level-set of a smooth hence continuous hence measurable function on $\mathbb{R}^n$. So the manifold is a Borel subset of $\mathbb{R}^n$. You can extend $\pi$ to all $\pi^*$ on all of $\mathbb{R}^n$ by letting $\pi^*(A)=\pi(A\cap\mathcal{M})$. If $\mathcal{C}$ itself is measurable, the restriction of $\pi^*$ to $\mathcal{B}(\mathcal{C})$ does the job. If $C$ is not measurable, you can take the measure to be $\pi^{**}$ with $\pi^{**}(A)$ being the $\pi^*$-outer measure of $A\cap C$. That this still defines a suitable measure is shown, for example, in 214A of the second volume of Fremlin's treatise.

$\endgroup$
11
  • $\begingroup$ Thank you! If you don't mind, can you give some more references about $\pi^*(A):= \pi(A\cap \mathcal{M})$? Does it have a name? $\endgroup$ Commented Nov 1, 2022 at 13:11
  • 1
    $\begingroup$ Fremlin uses the term "subspace measure." $\endgroup$ Commented Nov 1, 2022 at 13:18
  • 1
    $\begingroup$ Yes, the subspace measure is defined on the trace $\sigma$-algebra. $\endgroup$ Commented Nov 1, 2022 at 13:25
  • 1
    $\begingroup$ $\{0\}$ is a closed set and closed sets are Borel as the completementbof open sets. $\endgroup$ Commented Nov 1, 2022 at 13:59
  • 1
    $\begingroup$ I'm not really familiar with Hausdorff measures; sorry. $\endgroup$ Commented Nov 1, 2022 at 14:52

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .