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Edit: answered in the first comment. This turned out to be really easy, as the orbits of an open $\sigma$-compact subgroup yield a partition of $X$ into open $\sigma$-compact subsets.

Let $G$ be a locally compact Hausdorff topological group and consider the homogeneous space $X=G/H$ endowed with the quotient topology, where $H$ is a closed subgroup of $G$. Let $\mu$ be a nonzero Borel regular $G$-invariant measure on $X$ (or, more generally, a nonzero Borel regular measure on $X$ such that the action of each $g\in G$ on $\mu$ equals the product of $\mu$ by a positive scalar $\chi(g)$). By Borel regular I mean that $\mu$ is a countably additive nonnegative measure on the Borel $\sigma$-algebra such that the measure of a Borel subset equals the infimum of the measures of open sets containing it, the measure of an open subset equals the supremum of the measures of the compact sets contained in it and the measure of compact subsets is finite.

Is it true that if $B$ is a Borel subset of $X$ with finite measure then $B$ is contained in a $\sigma$-compact subset of $X$? By the Borel regularity of $\mu$, it is sufficient to consider the case in which $B$ is open. The result is valid, for instance, if $X$ is paracompact: in this case, $X$ is a disjoint union $X=\bigcup_{i\in I}X_i$ of $\sigma$-compact open subsets $X_i$ and an open subset of $X$ with finite measure intersects only countably many $X_i$. This observation takes care of the case in which $G$ admits an open $\sigma$-compact subgroup $G_0$ such that $G_0H$ is a subgroup of $G$ (namely, the projections onto $X$ of the left cosets of $G_0H$ yield a disjoint decomposition of $X$ into open $\sigma$-compact subsets). This includes the case in which $H$ is normal and the case in which $G$ is locally connected (in which case one can take $G_0$ to be the connected component of $1$, which is normal).

This result would be necessary to conclude the proof of Corollary 2.7.15 on Federer's book on geometric measure theory. If this result is false, then that Corollary will also be false.

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    $\begingroup$ It's known that $G\to G/H$ is an open map (Prop.6.9 in Stroppel's book). Hence, if $L$ is any open $\sigma$-compact subgroup, the $L$-orbits form a partition of $G/H$ into clopen $\sigma$-compact subsets. $\endgroup$
    – YCor
    Mar 4, 2019 at 1:43
  • $\begingroup$ But the result is false: for instance, fix an uncountable discrete group $\Gamma$: then in $\Gamma\times\mathbf{R}$, the subset $\Gamma\times\{0\}$ has measure zero but is not contained in any $\sigma$-compact subgroup. What follows from my previous comment is that any subset of finite measure is contained, up to measure zero, in a $\sigma$-compact (open) subset. $\endgroup$
    – YCor
    Mar 4, 2019 at 1:48
  • $\begingroup$ Thanks, your first comment proves the result. What you wrote in your second comment is not a counterexample, as the product measure is not Borel regular. The measure of $\Gamma\times\{0\}$ is zero, but every open subset of $\Gamma\times\mathbb{R}$ containing $\Gamma\times\{0\}$ has infinite measure, so outer regularity fails. $\endgroup$ Mar 4, 2019 at 2:43

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