I am studying a proof of the existence of Haar measure on locally compact groups.
http://www.albanyconsort.com/HaarMeasure/HaarMeasure.pdf
In this proof (at the top of page 7) when proving finite additivity of a content (which is like a measure but is only defined over compact sets) we separate the disjoint compact sets $K_1$ and $K_2$ by an open set $O^{-1}$ such that $K_1O^{-1}$ and $K_2O^{-1}$ are still disjoint. I am unsure why we are allowed to do this.
There is a general theorem where if $K \subseteq U$ where $K$ compact and $U$ open then there is an open set $V$ such that $KV \subseteq U$. So if the space is Hausdorff the compact sets are closed and this can be applied. But I don't think we are assuming Hausdorff. Regularity might be useful as well, but it is not quite the same. And unless we can generate some closed sets it cannot be applied.
Can two compact sets always be separated by an open set in a locally compact topological group? I see why they can is the group is Hausdorff. But I also believe the Haar measure exists on a non Hausdorff group.
Can someone either explain to me why the sets can be separated, or direct me to another proof of why Haar measure is additive in a non Hausdorff Group? And then I can patch it in.
Maybe you take the Kolmogorov Quotient of your group which is Hausdorff, prove Haar measure on it, and then use this to define a measure on the non Kolmogorovised group?
As far as I can see, the text you are referring to defines a Haar measure as a special kind of Radon measure, which is usually only defined for Hausdorff spaces.
In any case, there are simple examples of locally compact non-Hausdorff groups where Haar measure doesn't make sense. To be precise: there are locally compact topological groups where the only locally finite, translation-invariant measure in which compact sets are measurable is the zero measure.
The simplest example I can think of is the $(\mathbb{Z}, +)$ with the indiscrete topology. Since the smallest neighbourhood of $0$ is $\mathbb{Z}$, the measure has to be finite in order to be locally finite. On the other hand, any singleton is compact, and by translation-invariance each has the same measure $m$. But then $\sum_{n \in \mathbb{Z}} m$ is finite, therefore $m = 0$.
Addendum: If you only require the open sets to be measurable, in stead of compact sets, you could use the Haar measure of the Kolmogorov quotient, but there would not necessarily be a way to make any proper subset of an indiscrete subgroup measurable.