A book I'm reading discusses the proof of the following statement:
Let $S$ be a similarity transformation of scale factor $\lambda > 0$. If $F \subset \mathbb{R}^n$, then $$\mathcal{H}^s(S(F)) = \lambda \mathcal{H}^s(F)$$
The proof follows:
If $\{U_i\}$ is a $\delta$-cover of $F$ then ${S(U_i)}$ is a $\lambda\delta$-cover of $S(F)$, so $$\sum |S(U_i)|^s = \lambda^s \sum|U_i|^s$$ so $$\mathcal{H}_{\lambda \delta}^s(S(F)) \leq \lambda \mathcal{H}_\delta^s(F)$$Letting $\delta \rightarrow 0$ gives the $\leq$ direction. Replacing $S$ by $S^{-1}$ and so $\lambda$ by $1/\lambda$ and $F$ by $S(F)$ gives the opposite inequality.
I dont understand why from $\sum |S(U_i)|^s = \lambda^s \sum|U_i|^s$ we can't conclude $\mathcal{H}_{\lambda \delta}^s(S(F)) = \lambda \mathcal{H}_\delta^s(F)$
Even if I wanted to follow the instructions in the last sentence, I'm getting a little mixed up where everything should go in the argument marked in bold.
By definition $H^s_\delta$ reads: $$ H_\delta^s(F) := \inf\{ \sum |U_i|^s: U_i \text{ is a } \delta\text{-cover of } F \}. $$
The equality $\sum |S(U_i)|^s = \lambda^s \sum |U_i|^s$ gives $$ \inf\{ \sum |S(U_i)|^s: U_i \text{ is a } \delta\text{-cover of } F \} = \inf\{ \lambda^s \sum |U_i|^s: U_i \text{ is a } \delta\text{-cover of } F \} = \lambda^sH^s_\delta(F). $$ Notice that the infima on the lefthand side is a priori not over all covers $\lambda\delta$-covers of $S(F)$ just the ones who can be written as an image of $S$. From that observation we derive $$H^s_{\lambda\delta}(S(F)) = \inf\{ \sum |V_i|^s: V_i \text{ is a } \lambda\delta\text{-cover of } S(F) \} \leq \inf\{ \sum |S(U_i)|^s: U_i \text{ is a } \delta\text{-cover of } F \}.$$
This inequalities put together give $$ H^s_{\lambda\delta}(S(F)) \leq \lambda^sH^s_\delta(F). $$ Letting $\delta$ tend to $0$ you get $$ H^s(S(F)) \leq \lambda^sH^s(F). $$ Applying above results also to $S^{-1}$ yields the inequalitiy chain $$ H^s(S(F)) \leq \lambda^sH^s(F) = \lambda^sH^s(S^{-1}(S(F))) \leq \lambda^s \frac{1}{\lambda}^sH^s(S(F)) = H^s(S(F)) $$ and therefore $$ H^s(S(F)) = \lambda^sH^s(F). $$