Let $G$ be a real linear algebraic group (so it is locally compact and Hausdorff), equipped with a left-invariant Haar measure. Let $U$ be a Zariski open dense subset of $G$. I wonder how to show that $G-U$ has zero Haar measure?
My idea is that maybe we can first show that $G-U$ is contained in some lower dimensional manifold and conclude that it has zero Haar measure, but I don't know how to realize this more formally.
This follows directly from Sergei Ivanov’s rather stronger answer here. Density is not needed, only the fact that Haar measure is absolutely continuous with respect to the Lebesgue measure.