# Must a locally compact group be Hausdorff in order to possess a Haar measure?

Does the existence of (left) Haar measure on a locally compact topological group require that the group be Hausdorff?

• No, you can do the usual construction on the Kolmogorov quotient (which is locally compact Hausdorff) and pull the measure back on the original group. Commented Apr 28, 2010 at 3:39
• Perhaps I'm missing something obvious, but I'm not sure about "pull the measure back". Generally one can push measures forward though. Commented Dec 29, 2012 at 23:56
• The group on 2 elements with the indiscrete topology is locally compact, I think it has a Haar measure for any definition I can think of. Anyway, the question is unclear since "Haar measure" should be defined in a way that the question makes sense.
– YCor
Commented Mar 6, 2017 at 4:00
• @ToddTrimble since Borel subsets of a topological group are invariant by the closure of $\{1\}$, the definition of "pull the measure back" is clear. In general, if $f:X\to Y$ is a map and you have a $\sigma$-algebra and measure on $Y$, the pull-back of the $\sigma$-algebra is the set of $f^{-1}(B)$, $B$ in the $\sigma$-algebra of $Y$, and the measure is clearly defined. This construction is quite trivial and uninteresting but precisely is fine for this question.
– YCor
Commented Mar 6, 2017 at 4:02
• @YCor: You say that "the measure is clearly defined". Could you please make this part explicit for the slower ones of us ("lourds d'esprit")? I believe that the pullback measure should be $(f^* \mu) (B) = \mu (f (B))$, and it is not clear why $f(B)$ should be Borel when $B$ is Borel, and why this should give a measure. Or am I misunderstanding what "pullback measure" is supposed to mean here? Commented Apr 6, 2018 at 20:52

• @JasonDyer This comment is 7 years late: in Definition 5 of page 2 of that link one finds the notation $\mathcal{O}$ used to denote the collection of open sets, and on the 2nd to last line of page 3 it is thereby said that if $O$ is open and $K$ is a compact subset then $O \cap K^c$ is open. That is tantamount to knowing compact sets are closed. How does one know such closedness if the group is not assumed to be Hausdorff? Commented Mar 6, 2017 at 1:25
• @nfdc23 Isn't the property used in the paper actually equivalent (rather than just tantamount) to all compact sets being closed? It is asserted to work for all open sets $O$. So if we take $O$ to be the whole group, then the assertion is that $K$ compact implies $K^c$ open, right? If all compact sets are closed then the space is $T_1$. But topological groups are always completely regular. And comp. reg. + $T_1$ implies Hausdorff. Commented Jun 30, 2017 at 19:43
• @AlexM.: A $T_0$ topological group is $T_{3\frac{1}{2}}$, i.e. completely regular Hausdorff. Commented Feb 18, 2019 at 16:10