An outer measure $\mu^*$ is said to be regular if for every set $A \subset X$
$$\mu^\ast (A)=\inf\{\mu^*(E) : E\supset A \text{ is } \mu^\ast\text{-measurable} \}$$
To check that an outer measure is regular, we just have to check whether
$$\mu^\ast(A)\geq\inf\{\mu^*(E) : E\supset A \text{ is } \mu^\ast\text{-measurable} \}$$
since the other inequality follows from the outer measure axioms.
I have to find out whether the $s$-dimensional Hausdorff outer measure
For any $s \geq 0$ and $\delta\gt0$ we define the $\delta$-approximating $s$-dimensional Hausdorff outer measure,
$$\mathfrak{h}_{s,\delta}^\ast (A)= \alpha_s \inf\left\{\sum_{i=1}^\infty \text{diam}^s(A_i): A \subset \bigcup_{i=1}^\infty A_i, \text{diam}(A_i)\lt\delta\right\}$$
and the $s$-dimensional Hausdorff measure,
$$\mathfrak{h}_{s}^\ast(A)=\sup_{ \delta\gt0} \mathfrak{h}_{s,\delta}^\ast(A)= \lim_{\delta \downarrow 0} \mathfrak{h}_{s, \delta}^\ast(A)$$
Here $0\lt\alpha_s\lt\infty$ is chosen so that for $s \in \mathbb{N}$ the $s$-dimensional Hausdorff measure of the $s$-dimensional unit cube is one.
is regular. Since the Lebesgue outer measure is one of these measures, and it is regular, I'm trying to prove the Hausdorff outer measure is regular. So far I've got to
for any $A \subset X $ and $\delta\gt0$
$$\mathfrak{h}_s^\ast(A) \geq \inf\{\mathfrak{h}_{s,\delta}^\ast(E):A \subset E \text{ is } \mathfrak{h}_s^\ast\text{-measurable}\}$$
but I don't know how to show I'm allowed to swap my sup and my inf. If indeed I am.