Hausdorff measure is Borel regular

If ${\mathcal{H}}^s(A)=\mathrm{\infty }$, take $G=X$. If $A=\mathrm{\varnothing }$, take $G=\mathrm{\varnothing }$. If $s=0$ and the measure is finite, $A$ is finite, hence closed and Borel. In Euclidean space a finite set is a $G_{\delta }$, by intersecting its open $1\mathrm{/}k$-neighbourhoods.

In the remaining case $s>0$ and $M={\mathcal{H}}^s(A)<\mathrm{\infty }$, choose for each $k\ge 1$ a $2^{-k}$-cover $(U_{kj})$ of cost at most $M+2^{-k}$. Replacing each set by its closure preserves diameter: approximate two closure points by original points and use the triangle inequality. Set $G={\bigcap }_k{\bigcup }_j\overline{U_{kj}}$. This is Borel and contains $A$.

For fixed $\delta >0$ and all sufficiently large $k$, the $k$th closed cover also covers $G$ at scale $\delta$. Hence ${\mathcal{H}}_{\delta }^s(G)\le M+2^{-k}$, so ${\mathcal{H}}_{\delta }^s(G)\le M$. Take the supremum over $\delta$ and use monotonicity to get equality. Infimising Borel-superset measures gives the displayed identity: every such value is at least ${\mathcal{H}}^s(A)$ and this $G$ attains it.

For Euclidean $X$ in the finite positive-exponent case, enlarge $U_{kj}$ to the open neighbourhood $V_{kj}=\{x:d(x,U_{kj})<{\eta }_{kj}\}$ with $0<{\eta }_{kj}<2^{-k-2}$ and $(\mathrm{diam}{U}_{kj}+2{\eta }_{kj}{)}^s\le (\mathrm{diam}{U}_{kj}{)}^s+2^{-k-j-1}$. Continuity of the positive power at every nonnegative finite base supplies these choices, including singleton sets. Then $G={\bigcap }_k{\bigcup }_j{V}_{kj}$ is $G_{\delta }$, each covering diameter is at most $2^{1-k}$, and each cost is at most $M+2^{1-k}$. The same fixed-scale argument proves equality.