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 $\lambda$$F$, and $\lambda$ with $\frac{1}{\lambda}$ to get the other inequality.