I have this definition for the $s$-dimensional outer Hausdorff measure for some subset $A$ of a metric space $(X,d)$:
$$\mathcal H_*^s(A):=\lim_{\epsilon\to 0^+}\mathcal H_\epsilon^s(A)\tag1$$
where
$$\mathcal H_\epsilon^s(A):=\inf\left\{\sum_{k=0}^\infty[\operatorname{diam}(O_k)]^s:A\subset\bigcup_{k=0}^\infty O_k,\, O_k\in\mathcal T,\,\operatorname{diam}(O_k)<\epsilon\right\}\tag2$$
for $s>0$, where $\mathcal T$ is the induced topology in $X$ and
$$\operatorname{diam}(A):=\sup_{x,y\in A}d(x,y)\tag3$$
Then I need to show that for the case of $\Bbb R^n$ with the standard topology (and I assume the euclidean metric) that $(1)$ is equivalent to
$$\mathcal H_*^s(A):=\lim_{\epsilon\to 0^+}\inf\left\{\sum_{k=0}^\infty[\operatorname{diam}(A_k)]^s:A\subset\bigcup_{k=0}^\infty O_k,\, A_k\subset\Bbb R^n,\,\operatorname{diam}(A_k)\le\epsilon\right\}\tag4$$
What I did to show this equivalence was: because the $A_k$ are bounded then for each $\operatorname{diam}(A_k)=r_k\ge 0$ there is an open ball $O_k$ such that $\operatorname{diam}(O_k)=r_k+\delta_k$ for any arbitrarily small $\delta_k>0$ and $A_k\subset O_k$.
Then when $\mathcal H_*^s(A)=\infty$ then $(1)$ and $(4)$ coincide, and when it is finite then necessarily for enough small $\epsilon>0$ there are sequences $(A_k)$ such that $\sum_{k=0}^\infty[\operatorname{diam}(A_k)]^s<\infty$, then setting $\delta_k:=r_k/m$ for some $m\in\Bbb N$ we find that
$$ \left|\sum_{k=0}^\infty[\operatorname{diam}(A_k)]^s-\sum_{k=0}^\infty[\operatorname{diam}(O_k)]^s\right|=\left|\sum_{k=0}^\infty r_k^s-\sum_{k=0}^\infty(r_k+\delta_k)^s\right|\\=\left|\sum_{k=0}^\infty r_k^s\left(1-\left(\frac{m+1}m\right)^s\right)\right|=\left(\left(\frac{m+1}m\right)^s-1\right)\left|\sum_{k=0}^\infty r_k^s\right|\tag5 $$ Then as $m\to\infty$ the difference goes to zero, so for each $\epsilon>0$ the infima of $(1)$ and $(4)$ agree, so they agree in the limit and $(1)$ and $(4)$ are equivalent.
My questions:
It is my proof correct or it lack something?
The inequality $\le$ on $(4)$ play a role or it would be the same to put $<$ instead? I can't see a reason why this could make a difference.