This is taken from Pestov's book "Dynamics of Infinite-Dimensional Groups".
Let $\left( X, d, \mu \right)$ be a space with metric and Borel measure and let $Y \subseteq X$ be a closed subset. A measure $\nu$ on $Y$ is called induced from $X$, or else Hausdorff measure, if for every Borel (in the subspace topology of Y) subset $A \subseteq Y$ one has

\begin{equation} \nu(A) = \lim \limits_{\epsilon \to 0^+} \frac{ \mu(A_\epsilon) }{ \mu(Y_\epsilon) } \end{equation}

where for any subset $B \in X$ and any $\epsilon > 0$

\begin{equation} B_\epsilon := \left \{ x \in X \vert \exists \, b \in B : d(b,x) < \epsilon \right \} \end{equation}

I have never encountered this notion before and searching for Hausdorff measure turns up something quite different. Searching for induced measure turns up nothing...
I am trying to understand when the above is well-defined. Clearly we have to require $\mu(Y_\epsilon) < \infty$ for some $\epsilon > 0$. But why do we require $Y$ to be closed? Does the above limit always exist for all Borel subsets of $Y$?

Note that the most meaningful use is probably when $\mu(Y) = 0$. e.g $X = \mathbb{R}^2$ and $Y = [0,1] x {0}$.

  • $\begingroup$ I am not familiar with this but I suspect there may be undesirable anomalies in some cases when Y is not closed and a lot of interesting results when Y is closed. $\endgroup$ Jun 26, 2017 at 22:55

1 Answer 1


By continuity from above, the limit exists when $0<\mu(Y)<\infty$.

In the case $\mu(Y) = 0$ it may or may not be useful to note that $\nu(A) = \frac{M(A)}{M(Y)}$ when $A$ and $Y$ have Minkowski content $M(A)$ and $M(Y) > 0$, respectively, i.e. $$ M(A) = \lim_{\epsilon \to 0} \frac{\mu(A_\epsilon \setminus A)}\epsilon = \lim_{\epsilon \to 0} \frac{\mu(A_\epsilon)}\epsilon $$ (and the same thing for $Y$) exists. However, not every set has Minkowski content. In $\mathbb R^n$ there is a connection between rectifiability and Minkowski content. (An $m$-rectifiable set has $m$-dimensional Minkowski content equal to its $m$-dimensional Hausdorff measure. The Minkowski content that I defined above would be called $(n-1)$-dimensional in $\mathbb R^n$.) Both notions also exist in metric spaces, but I don't know about their connection.

  • $\begingroup$ I do not see how to get continuity in order to take the limit. I clarified the definition in the question. $\endgroup$
    – iolo
    Jun 26, 2017 at 18:21
  • $\begingroup$ @iolo Are you asking about the first part? We have $\mu(A_\epsilon) \to \mu(A)$ and $\mu(Y_\epsilon) \to \mu(Y)$. If $\mu(Y)>0$, then this gives convergence of the quotient $\endgroup$
    – user296355
    Jun 26, 2017 at 18:28
  • $\begingroup$ Yes but if $\mu(Y) = 0$ (say line segment in $\mathbb{R}^2$) the above construction still makes a lot sense $\endgroup$
    – iolo
    Jun 26, 2017 at 18:32
  • $\begingroup$ @iolo It does, and $\mu(Y)=0$ certainly is the more interesting case. The only thing that came to my mind there is what I wrote regarding Minkowski content. $\endgroup$
    – user296355
    Jun 26, 2017 at 18:36

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