# Scaling property proof of Hausdorff measure

I need to prove that $$\mathcal H^s(\lambda F) = \lambda^s\mathcal H^s(F)$$. Now my argument is as follows:

Let $$\{U_i\}$$ be a $$\delta$$-cover of $$F$$, then $$\{\lambda U_i\}$$ is a $$\lambda\delta$$-cover of $$\lambda F$$. So: \begin{align} \mathcal H^s_{\lambda\delta}(\lambda F) &= \inf\{\sum|\lambda U_i|^s:\{\lambda U_i\} \text{ is a } \lambda\delta-\text{cover of }\lambda F\} \newline &= \lambda^s\inf\{\sum|U_i|^s:\{U_i\} \text{ is a } \delta-\text{cover of }F\} \newline &=\lambda^s\mathcal H^s_\delta(F) \end{align}

Finally, taking $$\delta\rightarrow 0$$ both sides we get the required result. Is this argument correct?

I have found a similar proof of this in Falconer's book, which goes as follows:

$$\mathcal H ^s_{\lambda\delta}(\lambda F)\leq \sum|\lambda U_i|^s=\lambda^s\sum|U_i|^s\leq\mathcal H^s_\delta(F)$$

Then the other inequality is obtained by taking $$1/\lambda$$ instead of $$\lambda$$, and taking limits of delta on both sides. I can understand the first inequality by the definition of the Hausdorff measure, since it is the infimum of all such sums. However I cannot understand the second inequality.

you have: $$H^s_{\lambda\delta}(\lambda F) \leq \sum_{i} | \lambda U_i|^{s}=\lambda^{s}\sum_{i}|U_i|^{s}$$
The $$U_i$$'s are now a $$\delta$$ cover for $$F$$, therefore you may take the infimum over all possible such covers, to obtain: $$H^s_{\lambda\delta}(\lambda F) \leq\lambda^{s}H^s_{\delta}(F)$$.
Now send $$\delta$$ to $$0$$ you get the one of the inequalities. You may now replace the role of $$F$$ and $$\lambdaF$$, and $$\lambda$$ with $$\frac{1}{\lambda}$$ to get the other inequality.