No, but more out of laziness I must admit. I feared the computations would have been a mess and that maybe there was some neat clever way to do that. Seeing your answer, I was wrong!

Or at least, that's how it works for me

Thanks! I read through your answer and it is crystal clear, exactly what I was thinking, better explained. It's certainly possible that the reduced boundary has finite Hausdorff measure but the topological boundary doesn't, but not if they are equal! I think the issue is simply that when you read something in an article that doesn't seem right, you always assume you're not understanding something, especially if it's written by famous people. You just forget everyone can make a mistake

Thanks! I knew the book by Da Prato, but didn't know about the references therein. In the book by Kuksin and Shirikyan I don't see any particular focus on Stokes (just had a quick glampse though), does that mean they only treat the more general Navier-Stokes equations and the Stokes case can just be treated as a particular case?

@user378654 exactly, I get what they are doing but the removed segments are not the reduced boundary, as you say. I mean, I didn't go through the example in detail (that's why I didn't report it), what puzzles me is that they claim that the implication $P(\Omega)< +\infty$ + $\partial \Omega = \partial ^* \Omega \implies \mathcal{H}^{n-1} (\Omega)= P(\Omega)$ is false.

Right! Of course, it was quite simple indeed, thanks!

Just an idea: a way to look at it is that the operator norm of $A^{-1}$ is equal to the reciprocal of the smallest singular value of $A$. Since the flow changes volumes in a controlled way, the smallest singular value of $A=\nabla _x X$ cannot be "too small", or in the other directions you couldn't stretch enough to have an admissible volume.