2
$\begingroup$

Let $ S_1= ( [0,1], d_1 ) $ and $ S_2 = ( [0,1], d_2 ) $ be two metric spaces, where $ d_1 = |x - y|$ and $d_2 = (1/2^i) $ where binary expansion of x and y matches upto $ i^{th} $ coordinate. Let $ \mathbb{A}^1 $ be a family of subsets of $ S_1 $, and $ \mathbb{A}^2 = \{ [ \alpha ] : \alpha \in E^{( \ast ) }, E = \{0,1\} \} $ where $ E^{( \ast ) } $ is all finite strings, and $ [ \alpha ] $ is $ \alpha $ cylinder for $ E^{( \omega ) } $ which is set of all infinite strings. For $ S_1 $ metric, base function $ c_1 $ is usual distance function in $ \mathbb{R} $, but in $ S_2 $ we use base function $ c_2 $ as $ c_2 ( [\alpha] ) = \frac{1}{2^{|\alpha|}} $, where $ |\alpha| $ is length of string.

We define a Hausdorff measure separately for metric $ S_1 $ and $ S_2 $ using corresponding base function in the following manner: for each $ \epsilon > 0 $ let $ \mathbb{A}^i_{ \epsilon } = \{ A \in \mathbb{A}^i : diam(A) \leq \epsilon \} $ then $ \bar{\mathcal{M}}^i (E) = \lim_{ \epsilon \rightarrow 0 } \bar{\mathcal{M}_\epsilon^i } (E) $, Where $ \bar{\mathcal{M}_\epsilon^i } (E) = \inf \sum_{A_n \in \mathbb{A}_\epsilon^i } c_i (A_n ) $, where $ \{ A_n \} $ is countable cover of $ E $.

My question is, Can we have any relation between these two Hausdorff measures?

For reference: Gerald Edgar's book, Measure, Topology and Fractal Geometry, proposition 6.3.1 page number 177.

$\endgroup$

1 Answer 1

4
$\begingroup$

The two measures are within a constant factor of each other. In K. Falconer, Fractal Geometry: Mathematical Foundations and Applications, see the material on "net measure". In the 2nd edition, that is page 36

$\endgroup$
1
  • $\begingroup$ Thank you sir. Instead of $ S_1 \, and \, S_2 $ if we take $ (E^{\omega}, d_1) \, and \, (E^{\omega}, d_2) \, where \, E=\{0,1} \,, d_1 \, \& \, d_2 $ as above. Still we get same relation between hausdorff measure?. Since from exercise 4.2.2 we get addressing function is not inverse Lipschitz. Page no.123 your book MTFG. $\endgroup$
    – Uswadkar Prashant Vasantrao
    Commented Jun 16, 2015 at 5:07

You must log in to answer this question.

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