# Hausdorff Measure of a Compact segment in $\mathbb R$ has infinite hausdorff measure for $s < 1$

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?

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 $$\delta \rightarrow 0$$.