  Let $X\in\text{Alex}\,^n(-1)$ be an $n$-dimensional Alexandrov space with
curvature $\ge -1$. Let the $r$-scale $(k,\epsilon)$-singular set $\mathcal
S^k_{\epsilon,\,r}(X)$ be the collection of $x\in X$ so that $B_r(x)$ is not
$\epsilon r$-close to a ball in any splitting space $\mathbb R^{k+1}\times Z$.
We show that there exists $C(n,\epsilon)>0$ and $\beta(n,\epsilon)>0$,
independent of the volume, so that for any disjoint collection
$\big\{B_{r_i}(x_i):x_i\in \mathcal S_{\epsilon,\,\beta r_i}^k(X)\cap B_1,
\,r_i\le 1\big\}$, the packing estimate $\sum r_i^k\le C$ holds. Consequently,
we obtain the Hausdorff measure estimates $\mathcal H^k(\mathcal
S^k_\epsilon(X)\cap B_1)\le C$ and $\mathcal H^n\big(B_r (\mathcal
S^k_{\epsilon,\,r}(X))\cap B_1(p)\big)\leq C\,r^{n-k}$. This answers an open
question asked by Kapovitch and Lytchak. We also show that the $k$-singular set
$\mathcal S^k(X)=\underset{\epsilon>0}\cup\left(\underset{r>0}\cap\mathcal
S^k_{\epsilon,\,r}\right)$ is $k$-rectifiable and construct examples to show
that such a structure is sharp. For instance, in the $k=1$ case we can build
for any closed set $T\subseteq \mathbb S^1$ and $\epsilon>0$ a space
$Y\in\text{Alex}^3(0)$ with $\mathcal S^{1}_\epsilon(Y)=\phi(T)$, where
$\phi\colon\mathbb S^1\to Y$ is a bi-Lipschitz embedding. Taking $T$ to be a
Cantor set it gives rise to an example where the singular set is a
$1$-rectifiable, $1$-Cantor set with positive $1$-Hausdorff measure.
