# Last step in proof of Hausdorff outer measure $\mathcal{H}^{s,*} (cA) = c^s \mathcal{H}^{s,*}(A)$

The problem is just showing that the Hausdorff outer measure scales as follows:

$$\mathcal{H}^{s,*} (cA) = c^s \mathcal{H}^{s,*}(A)$$

For some covering $$\{E_k\}$$ of $$A$$, we get that

$$cA \subset \cup_i (c E_i)$$

$$\mathrm{diam} (cE_i) < c \delta$$

where $$\mathrm{diam} (S)$$ is the diameter of the set $$S$$.

Therefore,

$$H_\delta^{s,*}(cA) \leq \sum_{i=1}^\infty (\mathrm{diam} (cE_i))^s = c^s \sum_{i=1}^\infty (\mathrm{diam}(E_i))^s$$

Take $$\inf$$ on both sides over the countable coverings of $$A$$

$$H_\delta^{s,*}(cA) \leq c^s H_\delta^{s,*}(A) \xrightarrow[]{\delta \to0} H^{s,*}(cA) \leq c^s H^{s,*}(A)$$

I am suposed to use $$\frac{1}{c}A$$ to obtain the remaining inequality, but since I have to go through the same procedure, I end up with

$$H^{s,*}(\frac{1}{c}A) \leq \frac{1}{c^s}H^{s,*}(A)$$

From here, I can't get

$$H^{s,*}(cA)\geq c^s H^{s,*}(A)$$

$$H(A)=H(c^{-1}\cdot c\cdot A)\leq c^{-s}H(c\cdot A)$$, so $$c^{s}H(A)\leq H(c\cdot A)$$.