Talk:Hausdorff measure

recent edits

There is no problem with uncountable singletons (now, I'm too lazy to explain in details). Let's say this is just the usual way the definition is put in maths.

Now, I had to take it back due to some inaccuracies reintroduced in the text. First, the balls are a known special case. They do not disturb the dimension, the measure however is different. I believe you're able to check it for yourself.

The outer measure goes before the measure in the construction and my text takes it into account. We may want to add why the Borel sets are measureable (I mean it's a metric outer measure)

Generally, the language of "subsets" could be more appropriate than "spaces" (I mean e.g. a Brownian path is not a "self-contained space", it's embedded in R^d).

Feel free to do further tweaks, overall reverts are troublesome. Cheers Expert in topology (talk) 22:08, 29 June 2008 (UTC)

By the current definition, every set has (n>0)-dimensional Hausdorff measure zero since it can be covered by a possibly uncountable collection of singleton sets, each of which has diameter zero. The restriction that the diameter of each set should be positive appears to be essential. siℓℓy rabbit (talk) 22:22, 29 June 2008 (UTC)

Perhaps the collection needs to be countable? siℓℓy rabbit (talk) 22:25, 29 June 2008 (UTC)

(edit conflict) Covering by balls or sets it does not matter. Either way would work and the constant normalization is not standard anyway. But why do you insist on putting back the mistakes you have made and were deleted. Any set has Hausdorff dimension zero according to the definition you insist on. Your statement about paths is not quite precise or is misleading. You need some assumptions on the path. It is better in the begining to say number of points, rather than counting measure, which is more technical (and would be just fine further down). Oded (talk) 22:29, 29 June 2008 (UTC)

I think restricting to countable covers would work just as well as to positive radius, though I think the latter is somewhat preferable. Oded (talk) 22:29, 29 June 2008 (UTC)

If we choose to cover by sets rather than balls, then we need to restrict to countable covers. It is not always possible to make the diameter positive. Oded (talk) 22:33, 29 June 2008 (UTC)

Wow, I must have been blind. Of course, the definition is restricted to countable delta-covers. Then one pays no special attention to singletons, sorry for the buzz. You may want to look into Szpilrajn's note even if that's not the best source for our purposes. Keeping r>0 makes no harm, though. In fact, it is sometimes required (could you explain why you think it's not always possible?)

I insist however on making it with arbitrary sets. Balls do matter and give a slightly different object. Checking this is not that immediate but I still expect someone to do it before re-introducing the text. Perhaps the balls are used in some contexts, feel free to add a comment. For the time being I think we should use arbitrary sets as a verified standard.

I do not care too much about some wordings (e.g. "counting measure" vs "number of points"). Some others matter more. For example, I find using the Hausdorff measure to study "the size of fractals" quite restrictive (trivializing). After all, the Hausdorff measure is a basic tool in the geometric measure theory. I'd suggest, say, 'the structure of fractals' instead (I used yet another wording in the new edit). Also, I fail to see why a simple "rectifiable path" in R^n needs "more assumptions" than (or is less preferable to) "a metric space that is an imbededed path in R^n". In fact it needs *less* assumptions. The equality H^1=length holds for the Jordan curves (tweaked in the new version).

Further, I believe that the exposition about the outer measure should be given before we arrive at the measure. This is how it is usually done - it's meaningless to say "our measure is an outer measure" because the latter was there before by the very construction. Logically, the story goes the other way round - if the obtained outer measure is a metric outer measure then its restriction to measurable sets (=the measure) 'covers' the Borel sigma-field.

Please note that you simply reverted my edits without addressing the concerns. It's too easy. Please discuss or double check since it looks like some mistakes have been (re)introduced. Expert in topology (talk) 08:36, 30 June 2008 (UTC)

Possible error

I believe that the sentence

"Note that $\scriptstyle H_{\delta }^{d}(S)$ is monotone decreasing in δ ..."

should have the word "decreasing" replaced with "increasing." Johnfranks (talk) 16:55, 29 September 2009 (UTC)

I don't think so: the limit value for $\delta =0$ is larger: for example, the interval $[0,1]$ has infinite $H^{d}$-measure when $d<1$ but $H_{1}^{d}([0,1])$ is finite. --Bdmy (talk) 13:31, 30 September 2009 (UTC)

Inaccurate Statement

In the section that claims the Hausdorff measure is a scaling of the Lebesgue measure for $R^{d}$, I believe this is only true if we assume the space is equipped with the Euclidean norm.2620:0:10E2:15:60BC:6A74:2E45:C655 (talk) 18:00, 16 September 2019 (UTC)

Covering with open/closed sets/balls

https://en.wikipedia.org/w/index.php?title=Hausdorff_measure&type=revision&diff=938787569&oldid=916711923

I removed the claim that closed and open covers give us different $H_{\delta }^{s}$ values. The cited book (Federer 1969, §2.10.2) allows the covering sets to have size $\delta$, while our definition requires them to be strictly smaller. E.g. EoM https://www.encyclopediaofmath.org/index.php/Hausdorff_measure uses this definition too.

"However, we can require the covering sets to be open or closed,"

It is enough to show that for an O open set that is strictly smaller than a given $\delta$ there is a C closed set that contains it, and that is also strictly smaller than $\delta$, and vice versa. Which is true in any metric space.

"In $\mathbb {R} ^{n}$ restricting the covering sets to be balls may change the measures"

I don't have any examples. I feel that in the case of the Sierpiński triangle or similar cases, it may be true.

"but does not change the dimension of the measured sets."

It is enough to show $H_{\delta }^{s}(S)\leq B_{\delta }^{s}(S)$ and $B_{2\delta }^{s}(S)\leq 2^{s}H_{\delta }^{s}(S).$

I don't know if is it true in more general spaces, I guess it is not. BTW do we really need this "covering with balls" thing in this article? –– B. Macho (talk) 09:53, 2 February 2020 (UTC)