main obstacle towards extending Theorem \[smzpowTh9\] beyond metrizable $\sigma$-compact spaces. We found two non-metric situations in which an appropriate generalization of Lebesgue’s Covering Lemma holds. Here is the first of the two[^2]:

\[lebesguecov\] Let $(G,*)$ be a ${\sf T}_0$ topological group, let $\mathcal{U}$ be an open cover of $G$ and let $K$ be a nonempty compact subset of $G$. Then there is a neighborhood $N$ of the identity of $G$ such that for each $x\in K$ there is a $U\in\mathcal{U}$ with $x*N\subseteq U$.

Let $\mathcal{U}$ be an open cover of $G$. For each $x\in K$ choose a $U(x)\in\mathcal{U}$ with $x\in U(x)$. Then choose for each $x$ a neighborhood $N_x$ of $G$’s identity such that $x*{N_x}^2\subseteq U(x)$. Now $\{x*N_x:x\in K\}$ is an open cover of $K$. Since $K$ is compact this cover has a finite subset that covers $K$, say $\{x_1*N_{x_1},\cdots,x_k*N_{x_k}\}$. Define $N$ by $$N=N_{x_1}\cap\cdots\cap N_{x_k}.$$ Then $N$ is as required. For consider any $k\in K$. Choose $i$ so that $k\in x_i*N_{x_i}$. Now consider any $q\in k*N$. Fix $a\in N$ with $q = k*a$. Then we have $q\in x_i*N_{x_i}*a \subseteq x_i*N^2_{x_i} \subseteq U(x_i)$. Since $q$ was an arbitrary element of $k*N$ we find that $k*N\subseteq U(x_i)$. $\Box$

Here is the version 