This is probably obvious, but I'm having trouble with it. It is an exercise in Falconer:

Show that $\mathcal{H}^s([0,1]) = \infty$ if $s \in [0,1)$

I can see that this should loosely be the case: consider a dyadic partition of $[0,1]$, then $\sum_i 2^i \left(\frac 1 {2^i}\right)^s$ doesn't converge for $s < 1$. This however gets me an upper bound on Hausdorff measure. How does one proceed?


1 Answer 1


There is a covering of $[0,1]$ by sets with diameters $d_i<\delta$ such that $$\mathcal{H}^s_\delta([0,1]) \geqslant \left( \sum_i d_i^s\right) - 1$$ (the 1 is arbitrary and could be replaced by any positive constant).

But $d_i^s = d_i^{s-1}d_i\geqslant\delta^{s-1}d_i$, so $$\mathcal{H}^s_\delta([0,1]) \geqslant \delta^{s-1} \left( \sum_i d_i\right) - 1 \geqslant \delta^{s-1}- 1,$$ since the sets cover an interval of length one.

Now, just take the limit as $\delta \rightarrow 0$.


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