Let X be a compact Hausdorff topological group and let m be the Haar measure on X. Can we find a meager set in X whose complement is m-null? I can do it when X is separable but I don't know if there could be a non separable counterexample. The only non separable examples I know are products of separable groups so this holds there.


Thanks for your help!



Meager set of full measure
I switch to call $G$ your group because $X$ is a weird notation. The answer is: yes iff $G$ is infinite.


I assume you have a proof for $G$ infinite separable.


Next you can deal with the general case thanks to the following lemma: *if $G$ is an infinite compact group, then it has an infinite separable quotient.*


Indeed, if the lemma is OK, then just take such a quotient and pull back a meager subset of full measure.


To prove the lemma, let $(g\_n)$ be an injective sequence in $G$. Consider a finite-dimensional representation that is injective on $\{g\_1,\dots,g\_n\}$. Then its image is a separable quotient $G\_n$ of $G$. Let $G\_\infty$ be the closure of the image of $G$ in $\prod\_n G\_n$. Then $G\_\infty$ is separable, and is a quotient of $G$. Moreover $G\to G\_\infty$ is injective on $\{g\_n:n\ge 0\}$ and hence $G\_\infty$ is infinite. Thus the lemma is proved.


