Hausdorff Measure

**Dinkydoe** · September 22nd 2010, 05:13 AM

I hope anyone can explain this to me, who is familiar with the Hausdorff measure:

See: Hausdorff measure - Wikipedia, the free encyclopedia

Wikipedia says: $\mathcal{H}_{\delta}^d(S)$ is monotone decreasing in $\delta$ , and I agree. Since the larger $\delta$ , the more covers of U are permitted...

Then they conclude: Hence $\lim_{\delta\to 0}\mathcal{H}_{\delta}^d(S)$ exists. Why? Doesn't it only imply that $\lim_{\delta\to\infty}\mathcal{H}_{\delta}^d(S)$ exists?

I don't understand this at all.

**Opalg** · September 22nd 2010, 07:44 AM

Originally Posted by Dinkydoe

I hope anyone can explain this to me, who is familiar with the Hausdorff measure:

See: Hausdorff measure - Wikipedia, the free encyclopedia

Wikipedia says: $\mathcal{H}_{\delta}^d(S)$ is monotone decreasing in $\delta$ , and I agree. Since the larger $\delta$ , the more covers of U are permitted...

Then they conclude: Hence $\lim_{\delta\to 0}\mathcal{H}_{\delta}^d(S)$ exists. Why? Doesn't it only imply that $\lim_{\delta\to\infty}\mathcal{H}_{\delta}^d(S)$ exists?

I don't understand this at all.

If $H_{\delta}^d(S)$ decreases as $\delta$ increases, then it will increase as $\delta$ decreases. Therefore it will have a limit as $\delta\searrow0$ provided that we allow the possibility of the limit being infinity. In fact, although the Wikipedia page doesn't explicitly say that limits can be infinite, it certainly envisages that possibility. You can see that by scrolling down to the section headed "Relation with Hausdorff dimension", which includes the definition $\dim_{\text{Haus}}(S) = \sup\bigl(\{d\geqslant0:H^d(S) = \infty\}\cup\{0\}\bigr).$

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